In the given figure, $A B C D$ and $B P Q$ are straight lines. If $B P = B C$ and $D Q$ is parallel to $C P$ , prove that
(i) $C P = C D$
(ii) $D P$ bisects $∠ C D Q$

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Solution

As $B P = B C$
$\Rightarrow ∠ B P C = ∠ P C B$
As exterior angle is equal to sum of two opposite interior angles
$\Rightarrow ∠ B P C + ∠ P C B = 4 x$
$\Rightarrow ∠ B P C + ∠ B P C = 4 x$
$\Rightarrow 2 ∠ B P C = 4 x$
$\Rightarrow ∠ B P C = 2 x = ∠ P C B$
Now $D Q | | C P$
$∠ Q D C = ∠ P C B$ [Corresponding angles]
$\Rightarrow ∠ Q D C = 2 x$
As exterior angle is equal to sum of two opposite interior angles
$\Rightarrow ∠ P C B = ∠ C P D + ∠ P D C$
$\Rightarrow 2 x = x + ∠ P D C$
$\Rightarrow ∠ P D C = x$
As $∠ Q D C = 2 x$ and $∠ P D C = x$
$\Rightarrow D P$ bisects $∠ C D Q$ [Hence Proved]
As $∠ P D C = x$ and $∠ C P D$ and since sides opposite to equal angles are equal , hence
$C P = C D$ [Hence proved]

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Similar questions

In the given figure, if AP = BP and DP = CP, then which of the following are true?

In the given figure, AC = AE. Select the statements that are true.

In the given figure, AC = AE.

Show that :

(i) CP = EP

(ii) BP = DP.

A B

C D

P

C P = 2

D P = 6

A P = 3

B P

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