In the figure, AB is parallel to PRQ, T and P are the midpoints of CQ and CA respectively and ART is a straight line. Prove that TR : RA = 1 : 2. Solution Point Q is the midpoint of BC (midpoint theorem) QB : QC QC […]

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 […]

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 […]