Question

In the given figure, two chords AB and CD intersect each other at the point P. prove that:

(i) ΔAPC ∼ ΔDPB

(ii) AP.BP = CP.DP

Answer 1

Let us join CB.

(i) In ΔAPC and ΔDPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

ΔAPC ∼ ΔDPB (By AA similarity criterion)

(ii) We have already proved that

ΔAPC ∼ ΔDPB

We know that the corresponding sides of similar triangles are proportional.

`:. (AP)/(DP) = (PC)/(PB) = (CA)/(BD)`

`=>(AP)/(DP) = (PC)/(PB)`

∴ AP. PB = PC. DP