BdMO National Higher Secondary 2019/9

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samiul_samin
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BdMO National Higher Secondary 2019/9

Unread post by samiul_samin » Mon Mar 04, 2019 9:53 am

Let $ABCD$ is a convex quadrilateral.The internal angle bisectors of $\angle {BAC}$ and $\angle {BDC}$ meets at $P$.$\angle {APB}=\angle {CPD}$.Prove that $AB+BD=AC+CD$.

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Abdullah Al Tanzim
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Re: BdMO National Higher Secondary 2019/9

Unread post by Abdullah Al Tanzim » Thu Mar 28, 2019 9:14 pm

Let $B'$ be a point on the line $AB$ such that $AB'=AC$ and $C'$ be a point on the line $DC$ such that $DC'=BD$.
So, it suffices to prove that $BB'=CC'$.

$\triangle AB'P \cong \triangle ACP$ $ \Rightarrow \angle APB'=\angle APC$ $ \Rightarrow \angle BPB'=\angle APD$.

Similiarly, $\triangle DBP \cong \triangle DC'P \Rightarrow \angle CPC'=\angle APD $
So, $\angle BPB'=\angle CPC'$.

In $ \triangle BPB'$ and $ \triangle CPC',$
$\angle BPB'=\angle CPC'$
$BP=PC'$
$B'P=PC$
So, $\triangle BPB' \cong \triangle CPC'$
So, $BB'=CC' \Rightarrow AB+BD=AC+CD$
$(Q.E.D.)$ :D :D
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein

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samiul_samin
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Re: BdMO National Higher Secondary 2019/9

Unread post by samiul_samin » Wed Apr 10, 2019 10:18 am

No need to draw a figure?

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