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