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Cyclic Quadrilateral Properties | Ptolemy Theorem | Proof of AMBCID 11

cyclic quadrilateral ptolemy theorem

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 \frac{AC}{CD}=\frac{AB}{DM} ⇒ AB x CD = AC x DM       … (1)

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

Thus, we get \frac{AC}{BC}=\frac{AD}{BM} ⇒ AD x BC = AC x BM       … (2)

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

ABCD is cyclic quadrilateral ⇒ 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 \frac{AC}{BD}=\frac{AB \cdot AD + CB \cdot CD}{BA \cdot BC+DA \cdot DC}.

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 \frac{PA}{PB}=\frac{AD}{BC}=\frac{PD}{PC}.

From first two ratios, \frac{PA}{PB}=\frac{AD}{BC} ⇒ \frac{AD}{PA}=\frac{BC}{PB} 

Multiplying both sides by AB, we get

⇒ \frac{AB \cdot AD}{PA}=\frac{AB \cdot BC}{PB}                           … (1)    

From last two ratios, \frac{AD}{BC}=\frac{PD}{PC} ⇒ \frac{AD}{PD}=\frac{BC}{PC} 

Multiplying both sides by CD, we get

⇒ \frac{AD \cdot CD}{PD}=\frac{BC \cdot CD}{PC}                           … (2)

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

Then, we get \frac{PA}{PD}=\frac{AB}{CD}=\frac{PB}{PC}.

From first two ratios, \frac{PA}{PD}=\frac{AB}{CD} ⇒ \frac{AB}{PA}=\frac{CD}{PD} 

Multiplying both sides by AD, we get

⇒ \frac{AB \cdot AD}{PA}=\frac{CD \cdot AD}{PD}                           … (3)

From (1), (2), (3), we get \frac{AB \cdot AD}{PA}=\frac{AB \cdot BC}{PB}=\frac{BC \cdot CD}{PC}=\frac{CD \cdot AD}{PD}.

Since \frac{x}{y}=\frac{m}{n}=\frac{x+m}{y+n} ,

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

\frac{AB \cdot AD + BC \cdot CD}{PA+PC}=\frac{AB \cdot BC+CD \cdot AD}{PB+PD} ⇒ \frac{AB \cdot AD + BC \cdot CD}{AC}=\frac{AB \cdot BC+CD \cdot AD}{BD}

Thus, we get \frac{AC}{BD}=\frac{AB \cdot AD + CB \cdot CD}{BA \cdot BC+DA \cdot DC}

ABCD is cyclic quadrilateral ⇒ \frac{AC}{BD}=\frac{AB \cdot AD + CB \cdot CD}{BA \cdot BC+DA \cdot DC}

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