Question

Ans.
14. 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 * DP.
[3]
15.1​

Answer 1

in triangle BPA and triangle CPD

angle BPA = angle CPD    ( V.O.A)

angle BAC = angle CDB    ( 90 degree each )

therefore,

triangle PBA similar triangle CPD

AP/DP = PB/PC     (by bpt)

hence AP*PC=BP*pc