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
Solution
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
Is there an error in this question or solution?
Solution 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 Concept: Criteria for Similarity of Triangles.
