IGO 2016 Elementary/5

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Thamim Zahin
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IGO 2016 Elementary/5

Unread post by Thamim Zahin » Tue Jan 10, 2017 2:28 pm

Let $ABCD$ be a convex quadrilateral with these properties:
$\angle ADC = 135$ and $\angle ADB- \angle ABD = 2 \angle DAB = 4 \angle CBD$. If $BC$ $=$ $ \sqrt{2} CD$
prove that, $AB = BC + AD$.
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joydip
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Re: IGO 2016 Elementary/5

Unread post by joydip » Tue Jan 10, 2017 2:43 pm

$\angle DAB +\angle ABD +\angle ADB =180^\circ
\Rightarrow 2\angle CBD + \angle ABD + 4\angle CBD +\angle ABD =180^\circ $
So,$\angle CBD +\angle ABD +2\angle CBD = 90^\circ
\Rightarrow \angle ABC +\angle DAB =90^\circ $
So, $BC \perp AD$ .$E$ be the reflection of $D$ W.R.T $BC$ .Then $ E \in AD$.
$ \angle EBD = 2\angle CBD = \angle DAB$ . So , $\triangle EBD \sim \triangle EAB $ . As $BD = EB \Rightarrow AB = AE =AD + DE = AD + \sqrt{2} CD =AD +BC $ .
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