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# Cyclic Quadrilateral

If ABCD is a cyclic quadrilateral, then the sum of opposite angles is 180 degrees. It means, ∠A + ∠C = ∠B + ∠D = 180 degrees.

## Product of Diagonals : Ptolemy Theorem

In a cyclic quadrilateral, the sum of product of two pairs of opposite sides equals the product of two diagonals. This property of cyclic quadrilateral is known as **PTOLEMY THEOREM**.

If ABCD is a cyclic quadrilateral, then **AB x CD + AB x BC = AC x BD**.

**Proof:**

Take a point M on BD so that ∠ACB = ∠MCD.

As we know that the angles in same segment are equal. Then, ∠BAC = ∠BDC.

Now by AA Similarity, we have ∆ACB ~ ∆DCM.

Thus, we get ⇒ AB x CD = AC x DM … (1)

Now, ∠DAC = ∠DBC and ∠DCA = ∠BCM, then by AA similarity, we have ∆ACD ~ ∆BCM.

Thus, we get ⇒ AD x BC = AC x BM … (2)

From (1) + (2), we get **AB x CD + AD x BC = AC x BD**.

## Ratio of Diagonals

In a cyclic quadrilateral, the ratio of the diagonals equals the ratio of the sum of products of sides that share the diagonal’s end points.

If ABCD is a cyclic quadrilateral, then .

**Proof:**

Let ABCD is a cyclic quadrilateral whose diagonals intersect at P.

As we know that the angles in same segment are equal. Then, ∠BAC = ∠BDC and ∠ADB = ∠ACB.

From vertically opposite angles, ∠APB = ∠CPD and ∠APD = ∠BPC.

Now, by AA similarity, we have ∆PAD ~ ∆PBC.

Then, we get .

From first two ratios, ⇒

Multiplying both sides by AB, we get

⇒ … (1)

From last two ratios, ⇒

Multiplying both sides by CD, we get

⇒ … (2)

Now again, by AA similarity, we have ∆PAB ~ ∆PDC.

Then, we get .

From first two ratios, ⇒

Multiplying both sides by AD, we get

⇒ … (3)

From (1), (2), (3), we get .

Since ,

Adding (1st and 3rd Ratios) and (2nd and 4th Ratios)

⇒

Thus, we get

## Important Questions And Answer

**Ques 1** : In a triangle ABC, AB = 7, AC = 8, BC = 9. A point D is on its circumcircle such that AD bisects angle BAC. Then the value of AD/CD is

**Ans : 5/3**

**Ques 2** : E is a point on side AD of rectangle ABCD, so that DE = 6, while DA = 8, and DC = 6. If CE extended meets the circumcircle of the rectangle at F, find the measure of chord D.

**Ans : **

**Ques 3** : On side AB of square ABCD, right triangle ABF, with hypotenuse AB, is drawn externally to the square. If AF = 6 and BF = 8 find EF, where E is the point of intersection of the diagonals of the square.

**Ans : **

**Ques 4** : Point P on side AB of right triangle ABC is placed so that BP = PA = 2. Point Q is on hypotenuse AC so that PQ is perpendicular to AC. If CB = 3, find the measure of BQ, using Ptolemy’s Theorem.

**Ans : **

**Ques 5** : If any circle passing through vertex A of parallelogram ABCD intersects sides AB, and AD at points P and R, respectively, and diagonal AC at point Q, prove that (AQ)(AC) = (AP)(AB) + (AR)(AD).

**Ans : **

**Ques 6** : Diagonals AC and BD of quadrilateral ABCD meet at E. If AE = 2, BE = 5, CE = 10, DE = 4, and BC = 15/2 find AB.

**Ans : **

**Ques 7** : If isosceles triangle ABC (AB = AC) is inscribed in a circle, and a point P is on BC. Prove that PA/(PB + PC) = AC/BC a constant rate for the given triangle.

**Ans : **