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Top 120 Geometry Concept Tips and Tricks For Competitive Exams JSTSE NTSE NSEJS SSC

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Contents

Top 120 Geometry Concept Tips and Tricks For Competitive Exams JSTSE NTSE NSEJS SSC

ANGLES

Concept: 1 ANGLE

An ANGLE is the amount of rotation from initial arm to final arm which share a common endpoint. The arms are known as the sides of the angle and the common point is known is the vertex of the angle.

Concept: 2 COMPLEMENTARY ANGLES

If the sum of any two angles is equal to 90 degree or one right angle, then both the angles are known as COMPLEMENTARY ANGLES to each other. If the two complementary angles are adjacent angles then their non-common arms are either sides of common side and they form a right angle. 

Concept: 3 SUPPLEMENTARY ANGLES

If the sum of any two angles is equal to 180 degree or two right angle, then both the angles are known as SUPPLEMENTARY ANGLES to each other. If the two supplementary angles are adjacent, then their non-common sides form a straight line. Such angles are also known as Linear Pair of Angles.

Concept: 4 VERTICALLY OPPOSITE ANGLES.

A pair of angles opposite to each other formed by two intersecting straight lines that form an X-like shape are called VERTICALLY OPPOSITE ANGLES. They are always equal.

Concept: 5 CORRESPONDING ANGLES POSTULATE

If two parallel lines are intersected by a transvarsal line, then their corresponding angles are equal. This is a CORRESPONDING ANGLES POSTULATE.

Concept: 6 ALTERNATE INTERIOR ANGLE THEOREM

If two parallel lines are intersected by a transversal, then their alternate angles are equal. This is known as ALTERNATE INTERIOR ANGLE THEOREM.

Concept: 7 ANGLES SUM PROPERTY OF TRIANGLE

The sum of all interior angles in a triangle is 180 degree. The sum of all exterior angles of the triangle is 360 degree. This is known as ANGLES SUM PROPERTY OF TRIANGLE.

Concept: 8 EXTERIOR ANGLES PROPERTY OF TRIANGLE

The exterior angle of a triangle is the sum of two opposite interior angles. This is known as EXTERIOR ANGLES PROPERTY OF TRIANGLE.

Concept: 9 BISECTOR OF INTERIOR ANGLES

If the bisectors of angles ABC and ACB of a triangle ABC meet at a point I, then angle BIC90 ^{0} + \frac{1}{2}(Angle BAC)

Concept: 10 BISECTOR OF EXTERIOR ANGLES

The sides AB and AC of a triangle ABC are produced to P and Q respectively. If the bisectors of angle PBC and QCB intersects at I then BIC90 ^{0} - \frac{1}{2}(Angle BAC).

Concept: 11 FIND ANGLE

In a triangle PQR, if PS is the bisector of angle QPR (where angle Q > angle R) and PT \perp  QR, then angle TPS = \frac{1}{2}(Angle Q-AngleR). If the angle Q < angle R, then TPS = \frac{1}{2}(Angle R-AngleQ).

Concept: 12 FIND ANGLE

The angles between internal bisector of one base angle and the external bisector of the other base angle of a triangle is equal to one half of the vertical angle. Thus, we have angle BTC = \frac{1}{2}(AngleA).

Concept: 13 FIND ANGLE

In n-point star sum of all the angles at its vertices is (n-4)\times180 ^{0}. Thus, in a five-point star, we have the sum of all angles A + B + C+ D+ E = (5-4)\times180 ^{0}=180 ^{0}.

Concept: 14 FIND ANGLE

In a quadrilateral ABCD, if AO and BO are the bisectors of angles A and B respectively, then

angle AOB = \frac{1}{2}(AngleC+AngleD).

Concept: 15 SUM OF ALL ANGLES IN POLYGON

In any n sides convex polygon, the sum of all angles is (n-2)\times180 ^{0} and the sum of all exterior angles (taken in one direction, either clockwise of counter clock wise) in any convex polygon is 360 ^{0}.

Concept: 16 IMPORTANT RESULT

Two internal angle bisector of any two angles of a triangle cannot be perpendicular. 

Concept: 17 IMPORTANT RESULT

A convex polygon cannot have more than three acute internal angles.

CONGRUENT TRIANGLES


Concept: 18 Side-Angle-SIde (SAS) CONGRUENCE

Two angles are congruent, if two sides and the included angle of first triangle are equal to the corresponding sides and the included angle of the other triangle. In the figure below, AB = PQ angle A = angle P and AC = PR. Thus, \Delta ABC \simeq \Delta PQR

Concept: 19 Angle-Side-Angle (ASA) CONGRUENCE

Two angles are congruent, if two angles and the included sides of first triangle are equal to the corresponding angles and the included side of the other triangle. In the figure, AB = PQ angle A = angle P and angle B = angle Q. Thus, \Delta ABC \simeq \Delta PQR

Concept: 20 Angle-Angle-Side (AAS) CONGRUENCE

Two angles are congruent, if two angles and any non-included sides of first triangle are equal to the corresponding angles and the side of the other triangle. This is same as ASA Congruency.

Concept: 21 Side-Side-Side (SSS) CONGRUENCE

Two angles are congruent, if the three sides of first triangle are equal to the corresponding three sides of the other triangle. In the figure, AB = PQ, AC = PR and BC = QR. Thus, \Delta ABC \simeq \Delta PQR

Concept: 22 Right Angle-Hypotenuse-Side (RHS) CONGRUENCE

Two angles are congruent, if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the other triangle. In the figure, angle B = angle Q = 90 degree, AC = PR and BC = QR. Thus, \Delta ABC \simeq \Delta PQR

ISOSCELES TRIANGLE

Concept: 23 PROPERTY

Angles opposite to two equal sides of a triangle are equal. It means the base angles are equal.

Concept: 24 PROPERTY

The altitude drawn from one vertex of a triangle bisects the opposite side. It means the altitude and the median from the same vertex are same.

Concept: 25 PROPERTY

The bisector of the vertical angle bisects the base of the isosceles triangle.

Concept: 26 PROPERTY

Line l is the bisector of angle A and b is any point on l. BP and BQ are perpendiculars from B to the sides of the angle A. Then BP = BQ or B is outdistance from the sides of the angle A. That means, each point on the angle bisector is equidistant from the sides of the angle.

TRIANGLE INEQUALITIES

Concept: 27 PROPERTY

If the two sides of a triangle are unequal, the longer sides has the greater angles opposite to it.

Concept: 28 PROPERTY

In a triangle, the greater angle has the longer side opposite to it.

Concept: 29 PROPERTY

The sum of any two sides of a triangle is greater than the third side.

Concept: 30 PROPERTY

The difference of any two sides of triangle is less than the third side.

Concept: 31 PROPERTY

Of all the line segments that can be drawn to a given line, from a point not laying on it the perpendicular line segment is the shortest. In the figure, PM < PN

Concept: 32 PROPERTY

The sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle. In the figure, AP + BQ + CR < AB + BC + CA

 

Concept: 33 IMPORTANT RESULT

In a quadrilateral PQRS, PQ is the longest side and RS is shortest side. Then, Angle R>AngleP and Angle S>AngleQ.

Concept: 34 IMPORTANT RESULT

In a quadrilateral PQRS, diagonals PR and QS intersect at O. Then

(i) PQ + QR + RS + SP > PR + QS

(ii)PQ + QR + RS + SP < 2(PR + QS).

Concept: 35 IMPORTANT RESULT

O is any point in the interior of triangle ABC, then

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + OB + OC

(iii) OA + OB + OC > \frac{1}{2} (AB + BC + CA)

Concept: 36 IMPORTANT RESULT

In any triangle, the sum of any two sides is greater than the twice median drawn to the third side.

Concept: 37 IMPORTANT RESULT

In a triangle with the sides length a, b, and c and the medians length ma, mb and mc\frac{3}{4}(a+b+c)<(m_{a}+m_{b}+m_{c})<(a+b+c)

AREA RELATED CONCEPTS

Concept: 38 IMPORTANT RESULT

If D is any point the side BC of a triangle ABC, then \frac{ar(ABD)}{ar(ADC)}=\frac{BD}{DC}

Concept: 39 IMPORTANT RESULT

In ABCD is any quadrilateral, the diagonals AC and BD intersects at M. Then, ar(AMD)\times ar(BMC) = ar(AMB)\times ar(DMC).

Concept: 40 IMPORTANT RESULT

D, E and F are the points on the sides BC, CA, AB respectively of triangle ABC, such that AD, BE and CF are concurrent at P, then

(i) \frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}=1    (ii) \frac{AP}{AD}+\frac{BP}{BE}+\frac{CP}{CF}=2   (iii) \frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}

Concept: 41 IMPORTANT RESULT

A triangle ABC is divided into six smaller triangles by lines drawn from the vertices through a common interior point as shown in the figure. Then, use the following relation to find the unknown area of the smaller triangle.

(i) \frac{ar(BOD)}{ar(COD)}=\frac{ar(ABD)}{ar(ACD)}=\frac{BD}{CD}    (ii) \frac{ar(AOE)}{ar(COE)}=\frac{ar(ABE)}{ar(BCE)}=\frac{AE}{CE}   (iii) \frac{ar(AOF)}{ar(BOF)}=\frac{ar(ACF)}{ar(BCF)}=\frac{AF}{BF}

MID POINTS OF SIDES

Concept: 42 MID POINT THEOREM

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and half of it. In a triangle ABC, D and E are the mid points of AB and AC, then DE || BC and

DE = \frac{1}{2} BC.

Concept: 43 IMPORTANT RESULT

The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Concept: 44 MEDIANS

In a triangle ABC, the mid-points of the sides BC CA and AB are D E and F respectively. The lines AD BE and CF are called medians of the triangle ABC, the point of concurrency of the three medians is called centroid. The centroid divides the medians in the ratio of 2:1 from the vertex to the side. The centroid is denoted by G. Then,

(i) \frac{AG}{GD}=\frac{BG}{EG}=\frac{CG}{GF}=\frac{2}{1}  (ii)AG=\frac{2}{3}ADBG=\frac{2}{3}BE , CG=\frac{2}{3}CF

(iii) ar(BGD)=ar(CGD)=ar(CGE)=ar(AGE)=ar(AFG)=ar(BFG)=\frac{1}{6}(arABC)

Concept: 45 IMPORTANT RESULT

The mid point of the hypotenuse of a right angle triangle is equidistant from all its vertices. In a triangle ABC with angle B = 90 degree and D is the mid point of its hypotenuse, then BD = \frac{1}{2} AC.

Concept: 46 IMPORTANT RESULT

The line segment joining the mid points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to the half of the difference of these sides. In the figure, ABCD is a trapezium and P and Q are the mid-points of the diagonals AC and BD, then PQ=\frac{1}{2}(AB-CD) for AB > CD and PQ=\frac{1}{2}(CD-AB) for AB<CD.

Concept: 47 IMPORTANT RESULT

The line segment joining the mid points of the non-parallel sides of a trapezium is parallel to each of the parallel sides and is equal to the half of the sum of the parallel sides of the trapezium. In the figure, ABCD is a trapezium and P and Q are the mid-points of the non-parallel sides AD and BC, then PQ=\frac{1}{2}(AB+CD)

BASIC PROPORTIONALITY THEOREM

Concept: 48 BASIC PROPORTIONALITY THEOREM or THALES THEOREM

If a line is drawn parallel to one side of a triangle intersecting the other two sides, at distinct points, then it divides the other two sides in the same ratio. In the figure, ABC is a triangle with DE || BC. Then,we have the following result      (i) \frac{AD}{DB}=\frac{AE}{EB}  (ii)\frac{DB}{AD}=\frac{EC}{AE}  (iii) \frac{AB}{AD}=\frac{AC}{AE}  (iv) \frac{AD}{AB}=\frac{AE}{AC}  (v) \frac{AB}{DB}=\frac{AC}{EC}  (vi) \frac{DB}{AB}=\frac{EC}{AC} .

Concept: 49 Converse of BASIC PROPORTIONALITY THEOREM or THALES THEOREM

If a line divides any two sides of a triangle in the same ratio then the line must be parallel to the third side. In the figure, ABC is a triangle. If \frac{AD}{DB}=\frac{AE}{EB}, then DE || BC.

Concept: 50 IMPORTANT RESULT

On the sides BC, CA, AB of a triangle ABC, points D, E F are taken in such a way that \frac{BD}{CD}=\frac{CE}{AE}=\frac{AF}{FB}=\frac{2}{1}. Then the area of the triangle XYZ determined by the lines AD, BE and CF is \frac{1}{7}the of the area of the triangle ABC. It means ar(\triangle XYZ)=\frac{1}{7}ar(\triangle ABC).

ANGLES BISECTOR THEOREM

Concept: 51 INTERNAL ANGLES BISECTOR THEOREM

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. In the figure, ABC is a triangle. If AD bisects the angle A, then (i) \frac{AB}{AC}=\frac{BD}{DC}  (ii) \frac{ar\triangle ABD}{ar\triangle ACD}=\frac{BD}{DC}=\frac{AB}{AC}

Concept: 52 EXTERNAL ANGLES BISECTOR THEOREM

The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. In the figure, ABC is a triangle. If AD bisects the exterior angle A and intersects BC produced in D, then  \frac{AB}{AC}=\frac{BD}{DC}.

Concept: 53 LENGTH OF ANGLE BISECTOR

(i) If the internal angle bisector of angle A meets the opposite side BC at D, then AD =\sqrt{AB \times AC-BD \times DC}

(ii) If the internal angle bisector of angle B meets the opposite side AC at E, then BE =\sqrt{BA \times BC-AE \times EC}

(iii) If the internal angle bisector of angle C meets the opposite side AB at F, then CF =\sqrt{CA \times CB-AF \times FC}

SIMILAR TRIANGLE

Concept: 54 Side-Side-Side (SSS) SIMILARITY

If in two triangles the sides of one triangle are proportional to those of the other, then the triangles are similar and hence the corresponding angles of the two triangles are equal. In the figure, ABC and PQR are two triangles such that \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}, then \triangle ABC \sim \triangle PQR and hence the angles A=P, B=Q,C=R.

Concept: 55 Angle-Angle-Angle (AAA or AA) SIMILARITY

If in two triangles the angles of one triangle are proportional to those of the other, then the triangles are similar and hence the sides opposite to those angles are proportional. In the figure, ABC and PQR are two triangles such that A=P, B=Q,C=R, then\triangle ABC \sim \triangle PQR and hence  \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}

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Concept: 56 Side-Angle-Side (SAS) SIMILARITY

If in two triangles, one angle of one triangle is equal to one angle of the other triangle and sides containing these angles are proportional, then two triangles are similar and hence remaining corresponding angles are equal and the sides opposite to those angles are in the same proportional. In the figure, ABC and PQR are two triangles such that angle A = angle P and \frac{AB}{PQ}=\frac{AC}{PR}, then \triangle ABC \sim \triangle PQR and hence the angles A=P, B=Q,C=R and \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}.

Concept: 57 AREA RATIO THEOREM FOR SIMILAR TRIANGLES

The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides. In the figure, ABC and PQR are two triangles such that \triangle ABC \sim \triangle PQR, it means angles A=P, B=Q,C=R and  \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}, then \frac{ar\triangle ABC}{ar\triangle PQR} = \left ( \frac{AB}{PQ} \right ) ^{2}=\left ( \frac{BC}{QR} \right ) ^{2}=\left ( \frac{AC}{PR} \right ) ^{2}.

Concept: 58 IMPORTANT RESULT

In the triangles ABC and PQR, if AD and PM are the medians, AX and PY are the altitudes AL and PK are the angle bisectors and Sand S2 be their semi-perimeters respectively, then \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AX}{PY}=\frac{AD}{PM}=\frac{AL}{PK}=\frac{AB+BC+CA}{PQ+QR+RP}= \frac{\sqrt{arABC}}{\sqrt{arPQR}}.

Concept: 59 IMPORTANT RESULT

There is a unique triangle with sides length as 4, 5 and 6 units whose sides lengths are consecutive integers and one of whose angles is twice the other.

RIGHT ANGLE RELATED THEOREM

Concept: 60 IMPORTANT RESULT

If a perpendicular AD is drawn from the right angled vertex A of a right angled triangle ABC to the hypotenuse BC then the triangles on both sides of the perpendicular are similar to the whole triangles and to each other and hence AB^{2}=BD \times BCAC^{2}=CD \times CB, and AD^{2}=DB \times DC.

Concept: 61 PYTHAGORAS THEOREM

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the figure, ABC is a right angled triangle with angle A  = 90 degree, then BC^{2}=AB^{2}+AC^{2}.

Concept: 62 ACUTE ANGLED TRIANGLE THEOREM

In a triangle ABC, if angle B < 90 degree and AD \perp  BC, then AC^{2}=AB^{2}+BC^{2}-2BD \times BC

Concept: 63 OBTUSE ANGLED TRIANGLE THEOREM

In a triangle ABC, if angle B > 90 degree and AD \perp  BC, then AC^{2}=AB^{2}+BC^{2}+2BD \times BC.

Concept: 64 IMPORTANT RESULT

In a triangle ABC, if AC^{2}<AB^{2}+BC^{2}, then the triangle ABC is an acute angled triangle with B is acute angle. If AC^{2}>AB^{2}+BC^{2}, then the triangle ABC is an obtuse angled triangle with B is obtuse angle.

APOLLONIUS THEOREM

Concept: 65 APOLLONIUS THEOREM

In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. This theorem is used to find the length of the median of any triangle. In the figure, ABC is a triangle, in which AD, BE and CF are the medians which bisects the sides BC, AC and AB respectively. Thus, according to this theorem,

(i) AB^{2}+AC^{2}=2(AD^{2}+BD^{2})

(ii) BA^{2}+BC^{2}=2(BE^{2}+AE^{2})

(iii) CA^{2}+CB^{2}=2(CF^{2}+AF^{2})

Concept: 66 IMPORTANT RESULT

In any triangle, the three times the sum of the squares of three sides of a triangle is equal to four times the sum of the squares of its medians. In the figure, ABC is a triangle, in which AD, BE and CF are the medians which bisects the sides BC, AC and AB respectively. Then, 3(AB^{2}+BC^{2}+CA^{2})=4(AD^{2}+BE^{2}+CF^{2}).

STEWART THEOREM

Concept: 67 STEWART THEOREM

In a triangle ABC, if D is a point on side BC such that BD = m and DC = n and AD = d, then a(mn+d^{2})=b^{2}m+c^{2}n \Rightarrow (man+dad)=(bmb+cnc). To remember this theorem, use the trick (man+dad)=(bmb+cnc) or A man (product of m, a and n) and his dad (product of d, a and d) put a bomb (product of b, m and b)in the sink (product of c, n and c).

Concept: 68 STEWART THEOREM FOR MEDIAN

In the STEWART THEOREM, if AD is the median then m=n=\frac{a}{2} and the median AD = ma. Thus, by the STEWART THEOREM, we get a(\frac{a^{4}}{4}+m _{a} ^{2})=\frac{b^{2}a}{2}+\frac{c^{2}a}{2} \Rightarrow (b^{2}+c^{2})=(2m _{a} ^{2}+\frac{a^{2}}{2}). This is the APOLLONIUS THEOREM and the median lengths are

(i) m _{a}=\frac{1}{2}\sqrt{2b^{2}+2c^{2}-a^{2}}

(ii) m _{b}=\frac{1}{2}\sqrt{2c^{2}+2a^{2}-b^{2}}

(iii) m _{c}=\frac{1}{2}\sqrt{2a^{2}+2b^{2}-c^{2}}

Concept: 69 STEWART THEOREM FOR ANGLE BISECTOR

In the STEWART THEOREM, if AD is the angle bisector then m=\frac{ca}{b+c} and n=\frac{ba}{b+c}, AD = ta. Thus, by the STEWART THEOREM, we get a(\frac{a^{2}bc}{(b+c)^{2}}+t _{a} ^{2})=\frac{b^{2}ca}{b+c}+\frac{c^{2}ba}{b+c} \Rightarrow \frac{a^{2}bc}{(b+c)^{2}}+t _{a} ^{2}=bc.

Thus, the length of the angle bisectors are

(i) t_{a}=\sqrt{ bc\left [ 1-\frac{a^{2}}{(b+c)^{2}} \right ]}

(ii) t_{b}=\sqrt{ ca\left [ 1-\frac{b^{2}}{(c+a)^{2}} \right ]}

(iii) t_{c}=\sqrt{ ab\left [ 1-\frac{c^{2}}{(a+b)^{2}} \right ]}

QUADRILATERAL

Concept: 70 PRALLELOGRAM

A quadrilateral whose both pair of opposite sides are parallel known as parallelogram. Some of the properties of a parallelogram are below.

1 A diagonal of a parallelogram divides it into two congruent triangles.

2.In a parallelogram, the opposite sides are equal.

3.Two opposite angles are equal.

4. The diagonals of a parallelogram bisects each other.

5. In a parallelogram, the bisectors of any two consecutive angles intersects at right angle.

6. The angle bisector of a parallelogram form a rectangle.

7. In a parallelogram, the sum of any two consecutive angles is 180 degree.

8. In a quadrilateral, if both opposite sides are equal then, it is a parallelogram.

9. In a quadrilateral, if both opposite angles are equal then it is a parallelogram.

10. If the diagonals of quadrilateral bisects each other then it is a parallelogram.

11. If one pair of opposite sides equal and parallel then it is a parallelogram.

Concept: 71 RECTANGLE

A parallelogram in which anyone angle is right angle is called as rectangle. Some of the properties of a rectangle are below.

1. opposite sides are parallel and equal.

2. Opposite angles are equal and each of 90 degree.

3. Diagonals are equal and bisects each other.

4. When a rectangle is inscribed in a circle the diameter of the circle is equal to the diagonal of the rectangle.

5.For a given parameter of rectangle, a square has the maximum area.

6. The quadrilateral formed by joining the intersection of the angle bisector of a parallelogram is a rectangle.

7. If P is any point in the plane of the rectangle ABCD, then PA2 + PC2 = PB2 + PD2.

Concept: 72 RHOMBUS

A parallelogram in which any two adjacent sides are equal or all sides are equal, then it is called rhombus. Some of the properties of a rhombus are below.

1. Opposite sides are parallel.

2. All sides are equal.

3. Diagonals are perpendicular bisects the opposite pair of angles.

4. Diagonals bisects the opposite pair of angles.

5. A parallelogram is rhombus  if its diagonals are perpendicular to each other.

6. Any parallelogram circumscribing a circle is a rhombus.

Concept: 73 SQUARE

A square is a rectangle whose all sides are equal or a rhombus whose all angles equal. Thus each square is a parallelogram, a rectangle and a rhombus. Some of the properties of a square are below.

1. All sides are equal.

2. Opposite pair of sides are equal.

3. Diagonals are equal and are perpendicular bisector to each other.

4. Diagonal of an inscribed square is equal to the diameter of the circumscribing circle.

5. Sides of a circumscribing square is equal to the diameter of the inscribed circle.

6. Angles formed by the diagonals and a side of the square is each equal to 45 degree.

Concept: 74 TRAPEZIUM

A quadrilateral whose one pair of sides is parallel known as trapezium.  Some of the properties of a square are below.

1. The line joining the mid points of the non-parallel sides is half the sum of the parallel sides.

2. If the non-parallel sides are equal then the diagonals will also be equal to each to other and converse is also true. The corresponding trapezium is called ISOSCELES TRAPEZIUM.

3. Diagonals intersects each other proportionally in the ratio of the lengths of the parallel sides. In the figure, ABCD is a trapezium in which AC and BD intersects each other at P, then \frac{AP}{PC}=\frac{BP}{PD}=\frac{AB}{DC}.

4. By joining the midpoints of the adjacent sides of the trapezium four similar triangles are obtained.

5. If a trapezium is inscribed in a circle then it is an isosceles trapezium with non-parallel sides.

6. In an isosceles trapezium base angles are equal and other two angles are also equal.

Concept: 75 ISOSCELES TRAPEZIUM

If ABCD is an isosceles trapezium and the diagonals intersects at P then the following results are true.

(i) Non-parallel sides are equal, AD = BC

(ii) AC = BD

(iii) AP = PB and PD = PC

(iv) PA \times PC = PB \times PD

(v) \frac{PA}{PC}=\frac{PB}{PD}

(vi) Angle PAB = Angle PBA = Angle PDC = Angle PCD

(vii) Angle DAB = Angle CBA and Angle ADC = Angle BCD

(viii) Angle PAD = Angle PBC and Angle ADB = Angle ACB

(ix)  AC^{2}=AD^{2}+AB \times CD

(x) If PM || AB || CD then \frac{1}{PM}=\frac{1}{AB}+\frac{1}{CD}

(xi) ABCD is a cyclic quadrilateral and then all the properties of the cyclic quadrilateral also applicable.

Concept: 76 KITE

In a kite two pair of adjacent sides equal. Some of the properties of the kites are given below.

(i) AB = BC and AD = CD

(ii) Diagonals intersects at right angle.

(iii) Longer diagonal is the perpendicular bisector of the shorter diagonal.

(iv) The quadrilateral formed by the mid-points of the adjacent the sides of the kite is rectangle.

(v) Area of the Kite = \frac{1}{2} (Product of the diagonals).

CONCURRENCY AND COLLINEARITY

Concept: 77 CEVIAN

A line segment joining a vertex of a triangle to any point on the opposite side or the point may be the extension of the opposite side also is called a Cevian. In the figure, AD, BE and CF are the cevians of the triangle ABC.

Concept: 78 CONCURRENT LINES

Three straight lines are said to be concurrent if all the three lines passes through a common point.

Concept: 79 Three points are said to be collinear, if they lie on a straight line. If A, B and C are the three points on a straight line in order, then AB, BC and AC are taken to be positive and BA, CB and CA are taken to be negative. Thus, AB + BC = AC and AB + BC + CA = 0.

Concept: 80 CARNOT’S THEOREM

Let D, E and F be points on sides BC, AC and AB respectively of triangle ABC. Then, the perpendiculars to the sides of the triangle at points D, E and F are concurrent if and only if BD^{2}+CE^{2}+AF^{2}=DC^{2}+EA^{2}+FB^{2}

Concept: 81 CEVA’S THEOREM

If points D, E and F are taken on the sides BC, CA and AB of the triangle ABC so that the lines AD, BE and CF are concurrent at a point P (Inside or outside), then \frac{BD}{DC} \times \frac{CE}{EA} \times\frac{AF}{FB} =1 or  BD \times CE \times AF=DC \times EA \times FB .

Concept: 82 VAN AUBEL THEOREM

If points D, E and F are taken on the sides BC, CA and AB of the triangle ABC so that the lines AD, BE and CF are concurrent at a point P,  then

(i) \frac{CP}{PF} = \frac{CE}{EA} +\frac{CD}{DB}

(ii) \frac{AP}{PD} = \frac{AF}{FB} +\frac{AE}{EC}

(iii) \frac{BP}{PE} = \frac{BD}{DC} +\frac{BF}{FA}

 

Concept: 83 CONVERSE OF CEVA’S THEOREM

If three cevians AD, BE and CF satisfy \frac{BD}{DC} \times \frac{CE}{EA} \times\frac{AF}{FB} =1, then the three cevians are concurrent.

Thus, the following results are true.

(i) The medians of a triangle are concurrent.

(ii) The altitudes of a triangle are concurrent.

(iii) The internal bisectors of the angles  of a triangle are concurrent.

(iv) The internal angle bisector of an angle of a triangle and the other two external bisectors are concurrent.

Concept: 84 MENELAUS THEOREM

If a transversal cuts (Internally or externally) the sides BC, CA and AB of a triangle at D, E and F respectively then \frac{BD}{DC} \times \frac{CE}{EA} \times\frac{AF}{FB} =-1

 

Concept: 85 IMPORTANT RESULT

A transversal cuts the sides AB, BC, CD, DA of quadrilateral at P, Q, R and S respectively, then \frac{AP}{PB} \times \frac{BQ}{QC} \times\frac{CR}{RD}\times\frac{DS}{SA} =1

Concept: 86 IMPORTANT RESULT

The tangents at the vertices of a triangle to its circumcircle meets the opposite sides in three collinear points. In the figure, ABC is a triangle. The tangent to its circumcircle at A  meets opposite side produce BC at D. The tangent to its circumcircle at B  meets opposite side produce CA at E. The tangent to its circumcircle at C  meets opposite side produce BA at F. Then, the points D, E and F are collinear.

CIRCLE

Concept: 87 PROPERTY

One and only one circle can be drawn so as to pass three non-collinear points.

Concept: 88 PROPERTY

The perpendicular drawn from the center of a circle to a chord of the circle bisects the chord. Conversely, the straight line joining the mid-points of chord of circle to the center is perpendicular to the chord.

Concept: 89 PROPERTY

Equal chords of a circle are equidistant from the center. Conversely, if two chords are equidistant from the center then they are equal.

Concept: 89 PROPERTY

In the same circle or in equal circles, equal chords cut off equal arcs and conversely.

Concept: 91 PROPERTY

Angle subtended by an arc of a circle at the center of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.

Concept: 92 PROPERTY

Angles in the same segment of a circle are equal and conversely.

Concept: 93 PROPERTY

Angle in a semi-circle is right angle. Angle in a segment smaller than a semi-circle is an obtuse angle and angle in a segment bigger than a semi-circle is an acute angle.

Concept: 94 PROPERTY

Radius drawn at point of contact of a tangent to the circle is perpendicular to the tangent.

Concept: 95 PROPERTY

From an external point we can draw two tangents to the circle. Both the tangents are equal in length.

Concept: 96 PROPERTY

If two chords of a circle bisects each other, they are diameters.

Concept: 97 ALTERNATE SEGMENT THEOREM

If through a point on a circle, a tangent and a chord be drawn the angle which the tangent makes with the chord is equal to the angle in the alternate segment.

Concept: 98 INTERSECTING CHORDS THEOREM

If two chords AB and CD of a circle intersect at P (Internally or externally) then, PA \times PB=PC \times PD

Concept: 99 SECANT THEOREM

If through a point outside a circle a tangent and chord be drawn, then the square of the length of the tangent is equal to the rectangle contained by the segments of the chord.

 

COMMON TANGENTS TO TWO CIRCLE

Concept: 100 NUMBER OF COMMON TANGENTS

Let two circles C1 and C2 with centers O1 and O2 with radii R and r respectively with R > r and the distance between their centers is d, then the number of the common tangents that can be drawn to them varies from zero to four in the same plane of the circle depending upon the relative positions of the circles.

Case 1: The circle C2 lies wholly within C1 and the two circles do not touch other means d<R-r

\Rightarrow Number of Common Tangent = 0

Case 2: The circle C2 lies wholly within C1 and touches it internally at a point P means d=R-r

\Rightarrow Number of Common Tangent = 1

Case 3: The circle C1 and C2 intersect each other at two points means R-r<d<R+r

\Rightarrow Number of Common Tangent = 2

Case 3: The circle C1 and C2 touch each other externally means d=R+r

\Rightarrow Number of Common Tangent = 3

Case 4: The circle C2 lies outside from the circle C1 and the two circles do not intersect each other means d>R+r

\Rightarrow Number of Common Tangent = 4

Concept: 101 DIRECT AND TRANSVERSE TANGENTS

The direct common tangents to two circle intersects each other at a point on the line joining the centers. This point divides the line joining the centers externally in the ratio of their radii.

The transverse common tangents to two circles also intersect each other at a point on the line joining the centers. This point divides the line joining the centers internally in the ratio of their radii.

In the figure, The circle C1 and C2 do not intersect. In this case, there are four common tangents, two direct common tangents AB and XY and two transverse common tangents KL and MN.

The two direct common tangents AB and XY intersect at S1. The point S1 divides O1O2 externally in the ratio R:r. It means \frac{O_{1}S_{1}}{S_{1}O_{2}}=\frac{R}{r}.

Similarly, the two transverse tangents KL and XY intersect at S2. The point S2 divides O1O2 internally in the ratio R:r. It means \frac{O_{1}S_{2}}{S_{2}O_{2}}=\frac{R}{r}.

Concept: 102 LENGTH OF DIRECT COMMON TANGENTS

In the given figure, O1A = R and O2B = r, O1O2 = d and AB = TD = Length of Direct Common Tangent.

Then,  T_{D}^{2}=\sqrt{d^{2}-(R-r)^{2}}

If two circles touche each other externally then d = R + r. Thus,  T_{D}^{2}=\sqrt{(R+r)^{2}-(R-r)^{2}}=2\sqrt{Rr}

Concept: 103 LENGTH OF TRANSVERSE COMMON TANGENTS

In the given figure, O1K = R and O2L = r, O1O2 = d and KL = TI = Length of Transverse Common Tangent.

Then,  T_{I}^{2}=\sqrt{d^{2}-(R+r)^{2}}

If two circles touche each other externally then d = R + r. Thus,  T_{I}^{2}=\sqrt{(R+r)^{2}-(R+r)^{2}}=0

CYCLIC AND TANGENTIAL

Concept: 104 CYCLIC QUADRILATERAL

A quadrilateral which has a circle passing through all its four vertices is called a CYCLIC QUADRILATERAL. The sum of each pair of opposite angles is 180 degree. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. In the figure, ABCD is a cyclic quadrilateral. Then,

 Angle A + Angle C = Angle B + Angle D = 180^{\circ} and AnglePBC=Angle ADC

Concept: 105 SIMSON WALLACE LINE

The feet L, M, N of the perpendiculars on the sides BC, CA, AB of any triangle ABC from any point X on its circumcircle of the triangle ABC are collinear. The line LMC is called the SIMSON WALLACE LINE.

Concept: 106 PTOLEMY THEOREM FOR CYCLIC QUADRILATERAL

In a cyclic quadrilateral, the product of the diagonals is equal to the sum of the product of the pair of opposite sides. In the figure, ABCD is a cyclic quadrilateral. Then, AB \times CD + AD \times BC = AC \times BD.

Concept: 107 PTOLEMY THEOREM FOR ANY QUADRILATERAL

For any quadrilateral, the product of the diagonals is less than or equal to the sum of the product of the pairs of opposite sides. Equality holds for cyclic quadrilateral.

In the figure, ABCD is a any quadrilateral. Then, (AC\times BD)^{2}}=(AB\times CD)^{2}}+(AD\times BC)^{2}}-2(AB\times CD)(AD\times BC)cos(A+C)

For cyclic quadrilateral, A + C = 180 degree \Rightarrow cos(A+ C) = cos(180) = –1

\Rightarrow (AC\times BD)^{2}}=(AB\times CD)^{2}}+(AD\times BC)^{2}}+2(AB\times CD)(AD\times BC)

\Rightarrow AC \times BD = AB \times CD + AD \times BC

Concept: 108 PITOT THEOREM

If a circle is inscribed in a quadrilateral ABCD, then the sum of opposite sides are equal. it means AB + CD = AD + BC.

Concept: 109 IMPORTANT RESULT

Triangle ABC is equilateral triangle. D is on AB and E is on AC such that DE is tangent to the incircle Then, \frac{AD}{DB}+\frac{AE}{EC}=1

AREAS

Concept: 110 AREA OF TRIANGLE

There are following formula to find the area of the triangle.

Area of Triangle = \frac{1}{2}\times Base \times Height

Area of Triangle = \frac{1}{2} Product of Two Sides \times Sin (Included Angle between the Two sides)

Area of Triangle = \sqrt{s(s-a)(s-b)(s-c)}, where s=\frac{a+b+c}{2}

Concept: 111 AREA OF QUADRILATERAL

There are following formula to find the area of the quadrilateral.

Area of Quadrilateral = \frac{1}{2} Product of Two Diagonals \times Sin (Included Angle between the Two Diagonals)

Area of Quadrilateral = \sqrt{(s-a)(s-b)(s-c)(s-d)-abcdCos^{2}(\frac{SumOfOppositeAngles}{2})}, where s=\frac{a+b+c+d}{2}

For cyclic quadrilateral, the sum of opposite angles = 180 degree. Then, Cos^{2}\frac{(SumOfOppositeAngles)}{2}=Cos^{2}\frac{(180)}{2}=Cos^{2}(90)=0

Thus, Area of Cyclic Quadrilateral = \sqrt{(s-a)(s-b)(s-c)(s-d)}.

Concept: 112 AREA OF REGULAR POLYGON

There are following formula to find the area of the regular polygon.

In terms of its side, Let the side of the regular polygon is and the number of its sides = n, then

Area of Regular Polygon = \frac{ns^{2}}{4}cot\frac{\pi}{n}

CENTERS OF TRIANGLE

Concept: 113 ORTHOCENTER (O)

The point of intersection of altitudes from the vertices to the opposite sides of a triangle is known as the orthocenter of the triangle.

Orthocenter of an obtuse angled triangle lies outside the triangle and behind the obtuse angle.

Orthocenter of an acute angled triangle lies inside the triangle.

Orthocenter of a right angled triangle is at the right angle vertex.

Concept: 114 CENTROID

In a triangle ABC, the mid-points of the sides BC CA and AB are D E and F respectively. The lines AD BE and CF are called medians of the triangle ABC, the point of concurrency of the three medians is called centroid.

The centroid divides the medians in the ratio of 2:1 from the vertex to the side. The centroid is denoted by G. Then,

(i) \frac{AG}{GD}=\frac{BG}{EG}=\frac{CG}{GF}=\frac{2}{1}  (ii)AG=\frac{2}{3}ADBG=\frac{2}{3}BE , CG=\frac{2}{3}CF

(iii) ar(BGD)=ar(CGD)=ar(CGE)=ar(AGE)=ar(AFG)=ar(BFG)=\frac{1}{6}(arABC)

Concept: 115 CIRCUMCENTER

The point of intersection of perpendicular bisectors of the sides of a triangle is known as the circumcenter of the triangle.

It is the center of the circle which passes through the vertices of the triangle and the circle is known as circumcircle. The radius of the circumcircle is known as circum radius R.

Circumcenter is equidistant from the vertices of the triangle.

Circumcenter of an obtuse angled triangle lies outside the triangle.

Circumcenter of an acute angled triangle lies inside the triangle.

Circumcenter of a right angled triangle is at the mid-point of the hypotenuse.

Circumradius R = \frac{abc}{4\Delta }, where a, b and c are the sides of the triangle and \Delta is the area of the triangle.

Concept: 116 INCENTER

The point of intersection of angle bisectors of a triangle is known as the incenter of the triangle.

It is the center of the circle touches the sides of the triangle and the circle is known as incircle. The radius of the incircle is known as inradius r.

Incenter is equidistant from the sides of the triangle.

Incenter of always lies inside in any triangle.

Inradius r = \frac{\Delta}{s}, where \Delta is the area of the triangle and s is the semiperimeter.

Concept: 117 NINE POINT CENTER (N)

A circle passing through nine fixed points which are

(i) Three mid-points of sides of a triangle

(ii) Three feet of its altitudes

(iii) Three mid-points the line segment joining the orthocenter and vertex.

Thus, the circle passing through these nine fixed points is known as Nine Point Circle and its center is known as nine Point Center.

Concept: 118 ONGC

In any triangle, the orthocenter (O), the nine point center (N), the centeroid (G) and the circumcenter (C) are collinear.

The nine point center (N) is the mid-point of the orthocenter (O) and the circumcenter (C).

The radius of the nine point circle is the half of the circumradius of the triangle.

The centeroid (G) divides the line segment joining the orthocenter (O) and circumcenter (C) internally in the ratio of 2 : 1.

COSINE FORMULA

Concept: 119 COSINE FORMULA FOR ANY TRIANGLE

If ABC is a triangle whose sides are BC = a, CA = b and AB = c, then 

(i) cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}

(ii) cosB=\frac{c^{2}+a^{2}-b^{2}}{2ca}

(iii) cosC=\frac{a^{2}+b^{2}-c^{2}}{2ab}

TO BE CONTINUE

Concept: 120 UNDER RESEARCH

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