Question

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

PLEASE ANSWER....WILL MARK AS BRAINLIEST... ​

Answer 1

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

$\implies$2∠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

$\implies$2x = x + ∠PDC

$\implies$∠PDC = x

As ∠QDC = 2x and ∠PDC = x

$\implies$DP bisects ∠CDQ

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

So, CP = CD

Hence proved.

Answer 2

Answer:

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