In figure, considering triangles BEP and CPD, prove that BP × PD = EP × PC.
Given: A ∆ABC in which BD ⊥ AC and CE ⊥ AB and BD and CE intersect at P.
To Prove: BP × PD = EP × PC
Proof: In ∆EPB and ∆DPC, we have
∠PEB = ∠PDC [Each equal to 90º]
∠EPB = ∠DPC [Vertically opposite angles]
Thus, by AA-criterion of similarity, we have
∆EPB ~ ∆DPC
⇒ BP × PD = EP × PC
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