Question

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × PD.

Answer 1

GIVEN Right triangles `Delta ABC and Delta DBC` are drawn on the same hypotenuse BC and on the same side of BC. Also, AC and BD intersect at P.
TO PROVE `APxxPC=BPxxPD`
PROOF In `Delta BAP and Delta CDP`, we have
`angle BAP=angle CDP=90^(@)`
`angle BAP= angle CPD` (ver. opp `angle`)
`. angle BAP~ angle CDP` [ by AA- similarity]
`:. (AP)/(DP)=(BP)/(CP)`
`rArr APxx CP= BP xx DP rArr APxxPC=BPxxPD`.
Hence, `APxxPC=BPxxPD`