Question

In the given figure, ABCD and BPQ are straight lines. If BP=BC and DQ is parallel to CP, prove that : (i)CP=CD and (ii)DP bisects CDQ.

Answer 1

As BP is equal to BC, angle BPC is equal to angle PCB. As the exterior angle is equal to the sum of the two opposite interior angles. Angle BPC + Angle PCB = 4x. Angle BPC + Angle BPC = 4x. 2 times angle BPC = 4x. Angle BPC = 2x = Angle PCB. Now Line DQ is parallel to Line CP. Angle QDC = Angle PCB, due to corresponding angles. Angle QDC = 2x. As the exterior angle is equal to the sum of the two opposite interior angles. Angle PCB = Angle CPD + Angle PDC. 2x = x + Angle PDC. Angle PDC = x. As Angle QDC = 2x and Angle PDC = x. Line DP bisects Angle CDQ. As angle PDC = x and angle CPD and since sides opposite to equal angles are equal, hence Line CP is equal to Line CD.

Answer 2

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