Russian Olympiad 1996
- Thamim Zahin
- Posts:98
- Joined:Wed Aug 03, 2016 5:42 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 $\angle FAC =\angle EDB$.
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Re: Russian Olympiad 1996
Simple angle chasing
A smile is the best way to get through a tough situation, even if it's a fake smile.
- Kazi_Zareer
- Posts:86
- Joined:Thu Aug 20, 2015 7:11 pm
- Location:Malibagh,Dhaka-1217
Re: Russian Olympiad 1996
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]
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]
We cannot solve our problems with the same thinking we used when we create them.