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

(i) Triangle ABD is similar to triangle PDC

(ii) AB^{2 }= 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^{2 }= BD x PD