Math, asked by santoshbal86, 1 year ago

Two right triangles ABC and DBC are drawn from the same hypotenuse BC and on the same side of BC. If AC and BD intersect each other at P, prove that AP*PC=BP*DP​

Answers

Answered by sagar777707
0

Answer:

it's too easy take both the sides common factor and use Pythagoras theorem

Answered by viji18net
1

Answer:

In ∆APB & ∆DPC,

∠A =  ∠D = 90°

∠APB = ∠DPC

[ vertically opposite angles]

∆APB  ~ ∆DPC

[By AA Similarity Criterion ]

AP/DP = PB/ PC

[corresponding sides of similar triangles are proportional]

AP × PC = DP × PB

Hence, proved.

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