Problem 779

Solution:

As usual, let us denote $\text{ACB=}$ and $BC=a, CA=b, AB=c$, where without loss of generality $a>b$ and $\alpha>\beta$. Let $F$ be the point of intersection of the lines $DE$ and $AB$, and $\varphi$ the angle between these lines. From the ratios $\frac{BD}{DC}=\frac{c}{b}$ and $\frac{CE}{EA}=\frac{a}{c}$ we easily find $BD=\frac{ac}{b+c}, DC=\frac{ab}{b+c}$, $CE=\frac{ab}{a+c}$ and $EA=\frac{bc}{a+c}$. Menelaus's theorem for the line $DE$ and the triangle $ABC$ gives $AF=\frac{bc}{a-b}$ and $FB=\frac{ac}{a-b}$.

Now, by the Law of Sines in the triangles $FEA$ and $FDB$ we have
$$\text{( - ) = FEA EFA = FA EA = bc a-b bc a+c = a+c a-b ( + ) = FDB DFB = FB DB = ac a-b ac b+c = b+c a-b}$$
from which we obtain $\sin \varphi=\sin (\alpha-\varphi)-\sin (\beta+\varphi)=2 \sin \frac{\alpha-\beta-2 \varphi}{2} \cos \frac{\alpha+\beta}{2}<$ $\sin (\alpha-\beta-2 \varphi)$. Hence $\varphi<\alpha-\beta-2 \varphi$, i.e. $3 \varphi<\alpha-\beta$.