Question

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Answer 1

Your Answer:

Given:-

• BP = PC
• DQ parallel to CP
• ABCD and BPQ are the Straight lines

To Prove:-

• PQ = CD
• CP = CD, if DP bisects angle CDQ

Solution:-

As, BP is equal to BC

$\tt \angle BPC = \angle PCB$

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

$\tt \angle BPC + \angle PCB = 4x$

$\tt \angle BPC + \angle BPC = 4x$

$\tt \Rightarrow 2 \angle BPC = 4x$

$\tt \Rightarrow \angle BPC = 2x = \angle PCB$

Now Line DQ is parallel to Line CP

$\tt \angle QDC = \angle PCB$

Due to corresponding angles.

$\tt \angle QDC = 2x$

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

$\tt \angle PCB = \angle CPD + \angle PDC$

$\tt 2x = x + \angle PDC$

$\tt \angle PDC = x$

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