Question

In the given Figure, AB and CD are two chords of a circle, intersecting each other at P such that AP = CP. Show that AB= CD. 

Answer 1

If two chords of a circle interest internally then the products of the lengths of segments are equal, then  

AP x BP= CP x DP ... ( 1) 

But, AP= CP (Given)    ....(2)

Then from ( 1) and (2), we have 

BP= DP     ......(3)

Adding (2) and (3), 

AP + BP= CP + DP 

⇒ AB = CD 

Hence Proved. 

Answer 2

In order to prove the desired result, we shall first prove that ΔPAD ∼ ΔPCB.

In triangles PAD and PCB, we have:
∠ PAD = ∠PCB    ...[Angles in the same segment of arc BD]
∠ APD = ∠ CPB   ...[Vertically opposite angles]
So, by AAA criterion of similarity, we have
ΔPAD ∼ ΔPCB.

⇒ `"PA"/"PC" = "PD"/"PB"`  ....[Corresponding sides of similar triangles are in the same ratio]

⇒ `"AP"/"CP" = "PD"/"PB"` 

⇒ 1 = `"PD"/"PB"            ...[ ∴ "AP" = "CP", "AP"/"CP" = 1]`

⇒  PB = PD
⇒  AP + PB = AP + PD   ....[ Adding AP on both sides ]
⇒  AP + PB = CP + PD   ...[ ∵AP = CP ] 
⇒ AB = CD.
Hence proved.