In the given figure BD and CE intersect at p considering triangles BEP and CPD prove that BP.CD=EP.PC
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Given:
In ∆ABC, BD ⊥ AC and CE ⊥ AB and BD & CE intersect at P.
To Prove:
BP.PD = EP.PC
Proof:
In ∆EPB & ∆DPC
∠PEB = ∠PDC As it equal to 90 degree
∠EPB = ∠DPC They are vertically opp. angles
By AA-criterion of similarity
∆EPB ~ ∆DPC
Hence: BP.PD = EP.PC
In ∆ABC, BD ⊥ AC and CE ⊥ AB and BD & CE intersect at P.
To Prove:
BP.PD = EP.PC
Proof:
In ∆EPB & ∆DPC
∠PEB = ∠PDC As it equal to 90 degree
∠EPB = ∠DPC They are vertically opp. angles
By AA-criterion of similarity
∆EPB ~ ∆DPC
Hence: BP.PD = EP.PC
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