1.

In the figure, ABCD and BPQ are lines. BP=BC and DQ is parallel to CP. Prove that:a) CP=CD b)DP bisects

Answer»

GiVen:-

  • BP = BC

\implies∠BPC = ∠PCB

AnSwer:-

As the exterior angle is equal to the sum of two opposite INTERIOR ANGLES.

\implies∠BPC + ∠PCB = 4x

\implies∠BPC + ∠BPC = 4x

\implies2∠BPC = 4x

\implies∠BPC = 2x = ∠PCB

Now,

As DQ || CP

∠QDC = ∠PCB (corresponding angles)

\implies∠QDC = 2x

Here the exterior angle is equal to the sum of two opposite interior angles.

So,

\implies∠PCB = ∠CPD + ∠PDC

\implies2x = x + ∠PDC

\implies∠PDC = x

As ∠QDC = 2x and ∠PDC = x

\impliesDP bisects ∠CDQ

Since ∠PDC = x and ∠CPD and SIDES opposite to equal angles are equal,

So, CP = CD

Hence proved.


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