[2 Mark Question Answer]

Question 12 Marks

$\text{ABCD}$ is a parallelogram. The bisector of $\angle BAD$ meets $\text{DC}$ at $P,$ and $\text{AD}$ is half of $\text{AB}.$

Prove that$: \text{BP}$ bisects $\angle ABC.$

Answer

Since $\text{PC} =\text {BC} ...(\text{AD}$ is half of $\text{AB}$ and $\text{BC} =\text {AD}$ and $\text{DC} = \text{AB})$
$∠CPB = ∠CBP$
But $∠CPB = ∠PBA ...($alternate angles $∵\text{DC} \| \text{AB})$
$⇒ ∠CBP = ∠PBA$
Therefore, $\text{BP}$ bisects $∠ABC.$

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Question 22 Marks

$\text{ABCD}$ is a parallelogram. $P$ and $Q$ are mid$-$points of $\text{AB}$ and $\text{CD}.$ Prove that $\text{APCQ}$ is also a parallelogram.

Answer

In quandrilateral $\text {APCQ} $
$\text{AP} \| \text{Q C} \ldots($ since $\text{AB} \| \text{CD}) $
$\text{AP}=\frac{1}{2}\text{AB} \ldots ...$ given$) $
$\text{CQ}=\frac{1}{2} \text{CD} \ldots ($given$) $
But $\text{AB}=\text{CD}$
$\Rightarrow \text{AP}=\text{CQ}$
Therefore, $\text{APCQ}$ is a parallelogram.

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MATHEMATICS [2 Mark Question Answer]

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