hatchguy wrote: Let $ABC$ be a triangle with $AB < BC$, and let $E$ and $F$ be points in $AC$ and $AB$ such that $BF = BC = CE$, both on the same halfplane as $A$ with respect to $BC$. Let $G$ be the intersection of $BE$ and $CF$. Let $H$ be a point in the parallel through $G$ to $AC$ such that $HG = AF$ (with $H$ and $C$ in opposite halfplanes with respect to $BG$). Show that $\angle EHG = \frac{\angle BAC}{2}$. I think there's a small typo here. It should be "Let $H$ be a point in the parallel through $G$ to $AB$" instead of "$G$ to $AC$".

No, the problem is exactly as I wrote it.

@hatchguy I had misread the problem then. Now, here goes my proof: Let $I$ be the incenter of $\triangle ABC$. Note that $FAIC$ is cyclic since $\angle AIC=90^{\circ}+\dfrac{\angle B}{2}$ and $\angle AFC=90^{\circ}-\dfrac{\angle B}{2}$. So, $\angle AFI=\angle ACI=\dfrac{\angle C}{2}$ and $\angle IAC=\angle IFC=\angle ICF=\dfrac{\angle A}{2}$ (since $IF=IC$). Now, $\angle EBA=\angle BEC-\angle A=90^{\circ}-\dfrac{C}{2}-\angle A$. So $\angle IBE=\angle IBA-\angle EBA=\dfrac{\angle A}{2}$. Moreover, $IB=IE \Longrightarrow \angle IEB=\dfrac{A}{2}=\angle ICG \Longrightarrow GEIC$ is cyclic $\Longrightarrow \angle EGI=\angle ECI=\dfrac{\angle C}{2}=\angle AFI$. Also, $\angle GEI=180^{\circ}-\angle BEI=180^{\circ}-\dfrac{\angle A}{2}=\angle FAI \Longrightarrow \triangle FAI \sim \triangle GEI$ and so: $\dfrac{AF}{AI}=\dfrac{GE}{EI}=\dfrac{GE}{BI} \Longrightarrow \dfrac{AF}{GE}=\dfrac{HG}{GE}=\dfrac{AI}{BI}$. But we also have $\angle HGE=\angle GEC=90^{\circ}+\dfrac{\angle C}{2}=\angle AIB$ $\Longrightarrow$ $\triangle HGE \sim \triangle AIB \Longrightarrow \angle EHG=\angle BAI=\dfrac{\angle A}{2} \text{ } \square$

Haha this is my problem Oficial Solution: Let $I_{A}$ be the excenter of $\triangle ABC$ at vertex $A$. Is easy prove that $BGCI_{A}$ is a parallelogram, and then $ GB=CI_{A}$ and $\angle HGE=\angle ACI_{A}$. By Menelao's to $\triangle ABE$ and transversal $F-G-C$ : \[\frac{AF}{FB}\cdot\frac{BG}{GE}\cdot\frac{EC}{CA}=1\] But $FB=CE$, $GB=CI_{A}$ and $FA=HG$. Hence, $\displaystyle\frac{AC}{CI_{A}}=\frac{HG}{GE}$, implying $\triangle ACI_{A}\sim\triangle HGE$. And this similarity allows us to conclude $\displaystyle\angle EHG=\angle CAI_{A}=\frac{\angle BAC}{2}$.

Indeed, this problem is due to our friend Eduardo Aguilar from El Salvador, a.k.a. Concyclicboy!

What about a trigo solution.

panamath wrote: What about a trigo solution using $sen2\theta$... I'd really like to see it, I've been trying but I've failed. A trigo solution : $BG=\frac {a Cos (\frac {B}{2})}{Cos (\frac {A}{2})},BF=2a Sin (\frac {C}{2})$ Easy calculation gives $GE=\frac {(c-a)Sin (\frac {A}{2})}{Cos(\frac {C-A}{2})}$.So as $c-a=HG$, hence done.