two triangles bac and bad right angled at a and d are drawn on same base bc . if ac and bd intersect at p . prove that ap * pc=bp*pd

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Consider tr.ABP and tr.DCP,

angleA = angleD (90)

angleAPB =angleDPC (vertically opposite angles)

Therefore, tr.ABP is similar to tr.DCP(A.A)

Therefore AP/PD = BP/PC

By cross multiplying we get,

AP*PC = BP*PD (Hence Proved)

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