Russian Olympiad 1996

For discussing Olympiad level Geometry Problems
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Thamim Zahin
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Russian Olympiad 1996

Unread post by Thamim Zahin » Wed Aug 03, 2016 7:06 pm

Points $E$ and $F$ are on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$). It is known that $\angle BAE =\angle CDF$ and $\angle EAF =\angle FDE$. Prove that $\angle FAC =\angle EDB$.

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Raiyan Jamil
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Re: Russian Olympiad 1996

Unread post by Raiyan Jamil » Wed Aug 03, 2016 9:30 pm

Simple angle chasing
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Kazi_Zareer
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Re: Russian Olympiad 1996

Unread post by Kazi_Zareer » Thu Aug 04, 2016 10:22 am

Since, $\angle EAF = \angle FDE $ so $ADFE$ is cyclic.
So, $\angle FDA + \angle AEF = 180 $ degrees .....(1)

If we can prove that $ ADCB$ is a cyclic, then it'll be done.

Now, $\angle ABC + \angle ADC $= $(\angle FEA - \angle EAB) $ + $ (\angle FDA + \angle FDC)$ = $(\angle FDA + \angle AEF) +(\angle CDF - \angle BAE)$ = $180 $ degree [from (1) we get, $\angle FDA + \angle AEF = 180 $ degrees and $\angle BAE = \angle CDF$, as given]
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