Problems involving properties of similar triangles

In the diagram, ABCD is a rectangle and PC is perpendicualr to BD. Prove that

(i) Triangle ABD is similar to triangle PDC

(ii) AB2 = BD x PD

(i) 

∠ABD = ∠PDC (alternate ∠s, AB//OC)

∠DPC = ∠BAD = 90°(given)

Therefore by Angle Angle similarity, triangle ABD is similar to triangle PDC.

(ii)

AB/PD = BD/DC

AB x DC = BD x PD

Since, AB = DC

Therefore AB= BD x PD

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