Let the triangle $ABC $ with $m (\angle ABC)=60^{\circ} $ and $m (\angle BAC)=40^{\circ}$ . Points $D $ and $E $ are on the sides $(AB) $ and $(AC) $ such that $m (\angle DCB )=70^{\circ}$ and $m (\angle EBC)=40^{\circ}$ . $BE$ and $CD$ intersect in $F $ . Prove that $BC $ and $AF $ are perpendicular.
Problem
Source: Moldova TST 2018
Tags: geometry
Combinatorix_Fan
06.04.2018 17:06
Let $H_a$ be the foot of the height passing by $A$. We would like to show that $DC,BE,AH_a$ concur. In order to do this, we will use the trigonometric version of Ceva's theorem. We must prove: $$\frac{\sin{\angle{EBA}}}{\sin{\angle{EBC}}} \cdot \frac{\sin{\angle{DCB}}}{\sin{\angle{DCA}}} \cdot \frac{\sin{\angle{H_aAC}}}{\sin{\angle{H_aAB}}} = 1$$ After some simple angle chasing, this is equivalent to $$\frac{\sin{20}}{\sin{40}} \cdot \frac{\sin{70}}{\sin{10}} \cdot \frac{\sin{10}}{\sin{30}} = 1$$ $\iff \sin{20} \sin{70} = \sin{30} \sin{40}$ Using the fact that $\sin 70 = \cos 20$, this is equivalent to $\frac{1}{2} \sin{40} = \sin{30} \sin{40} \iff \sin{30} = \frac{1}{2}$, which is true.
WolfusA
06.04.2018 17:42
Nice solution. Of course measures of angles are in degrees, not radians.