SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND)

We have $\angle ADE=\angle B$ and $\angle ADF=\angle C$. So $AFDE$ is a cyclic



quadrilateral and $$\angle AFE=\angle ADE=\angle B\text{ and }\angle AEF=\angle ADF=\angle C,$$ Hence, $EF$ is parallel to $BC$ and $DE=DF$, so the triangle $FDE$ is isosceles.

Let $M$ be the midpoint of $EF$, $L=AD\cap EF$, and $H=DG\cap EF$. By Thales Theorem, we obtain $$\frac{LE}{LF}=\frac{DC}{BD}$$ Also, from $△FGH\sim △CGD$ it follows that $$\frac{HF}{DC}=\frac{GH}{GD}$$ From $△EGH\sim △BGD$, we get $$\frac{HE}{BD}=\frac{GH}{GD}$$ and from this $$\frac{HF}{HE}=\frac{DC}{DB}$$ then $H$ and $L$ are symmetric with center $M$, as we will see later and $\angle LDM=\angle HDM=\alpha$ with $$\alpha =90^{\circ }-\angle ADC=\frac{\angle C-\angle B}{2}\text{ if }C>B,$$ and $$\alpha =90^{\circ }-\angle ADB=\frac{\angle B-\angle C}{2}\text{ if }B>C.$$ Finally, $$\angle GDE=\angle ADE-\alpha =\angle ADF\text{ if }B>C,$$ and $$\angle GDE=\angle ADE+\alpha =\angle ADF\text{ if }B<C.$$ It remains to prove that $L$ and $H$ are symmetric with center $M$. To do so denote $$LF+LE=HF+HE=s,\text{ and then}$$ $$\frac{LE}{LF}=\frac{HF}{HE}$$ hence $$\frac{s}{LF}=\frac{s}{HE}\Rightarrow LF=HE\text{ and }LE=HF$$ that is, $HM=ML$, as we wanted to prove.