69*1. Slicing problem 3.

by Virgil Nicula, Jul 29, 2010, 6:14 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=314172&p=1692729#p1692729
Ulanbek_Kyzylorda KTL wrote:
Let $ ABC$ be a triangle and $ D\in (AB)$ , $ F\in (AC)$ be two points so that $ m(\angle BDC)=70^{\circ}$ , $ m(\angle CDF)=40^{\circ}$ , $ m(\angle BCD)=30^{\circ}$ and $ m(\angle ACD)=20^{\circ}$ . Find $ m(\angle DBF)$ .
Proof. Prove easily that $ \frac {BD}{BC}=\frac {FD}{FC}=\frac {1}{2\cos 20^{\circ}}$ . Hence the $ B$-bisector in $ \triangle CBD$ and the $ F$-bisector in $ \triangle CFD$ are concurrently in $ K\in (CD)$ . Denote $ L\in CD$ so that $ D\in (CL)$ and $ \frac {LD}{LC}=\frac {KD}{KC}$ . Observe that the points $ B$ , $ F$ belong to the circle with diameter $ [KL]$ . Denote $ x=m(\angle DBF)$ . Since $ \angle LBF\equiv\angle LKF$ and $ m(\angle LBF)=50^{\circ}+x$ , $ m(\angle LKF)=80^{\circ}$ obtain $ 50^{\circ}+x=80^{\circ}$ , i.e. $ x=30^{\circ}$ .
Virgil Nicula wrote:
Generalization. Let $ ABC$ be a triangle and $ F\in (AB)$ , $ E\in (AC)$ be two points so that $ BC\cdot EF=BF\cdot CE$, $ m(\angle AFE)=\alpha$ , $ m(\angle BCF)=\beta$ and $ m(\angle ACF)=\gamma$ . Prove that $ m(\angle ABE)=\frac {\alpha +\gamma -\beta}{2}$ .
This post has been edited 5 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:47 PM

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