In acute $\triangle ABC$, $D$ is midpoint of $BC$, $DE \perp AB$, $DF \perp AC$, and $CE = BF$. Prove "geometrically" that $AB = AC$.

Question

Really seemingly simple geometry problem:

In acute $\triangle ABC$, $D$ is midpoint of $BC$. $DE \perp AB$, $DF \perp AC$. $CE = BF$. Prove that $AB = AC$

I am looking for a more "geometrical" solution. We can apply cosine theorem on $\triangle BCF$ and $\triangle BCE$ to prove that $\cos B = \cos C$. But the computation is annoying. Is there a more pretty solution?

Answer 1

To show $AB=AC$, it's enough to show $EB = FC$ and $AE=AF$. To show the former, I suggest thinking about showing congruence of $\triangle EGB$ and $\triangle FGC$, where $G$ is the intersection of $EC$ and $FB$. Argue that $EG$ = $FG$ and $BG = CG$.

Maybe you can do something similar to show $AE=AF$ - the triangles $\triangle AEC$ and $\triangle AFB$ in particular.

Answer 2

Let $B'$ be the projection of $B$ on $AC$. Then $F$ is the mid point of $B'C$, and we have: $$ BF^2 =B'B^2 +B'F^2=4FD^2+FC^2 =3FD^2 + DC^2\ .$$ There is a similar relation for $CE^2$. From $DB=DC$ we obtain $FD=ED$. Then $\Delta FDC=\Delta EDB$, giving $\hat B=\hat C$ in $\Delta ABC$.