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 […]

Proving using angle properties of circles

In the figure, BCE and ADE are straight lines. Given that ∠BAD = ∠ACD, prove that  (i)CD bisects ∠ACE, (ii) BD = AD Solution: (i) Let ∠BAD and ∠ACD be x ∠DCB = 180° – x (∠s in the opposite segments) ∠ECB = 180° – (180° – x) ∠ECD = x Since , ∠ECD = ∠ACD = x Therefore CD bisects ∠ACE   […]