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.
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∆APB ~ ∆DPC (AA similarity)
AP/DP = BP/PC
⇒ AP × PC = BP × DP
Answered by
2
Answer:
AP×PC = BP × DP
Explanation:
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