Show that a point is a midpoint of a side of a triangle In $\Delta ABC$, the bisector of $\angle A$ intersects $BC$ at $D$. The perpendicular to $AD$ from $B$ intersects $AD$ at $E$. The line through $E$ parallel to $AC$ intersects $BC$ at $G$, and $AB$ at $H$.
Prove that $H$ is the mid-point of $AB$.
I have no idea on how to prove this question. Any hint would be much appreciated. 
 A: Trivial by picture:

$BEA$ is a right triangle and by Thales' theorem $\widehat{BGH}=\widehat{BCA}$, $\widehat{BHG}=\widehat{BAC}$.
Angle chasing gives $\widehat{HEA}=\widehat{EAC}=\widehat{HAE}$, hence $HA=HB=HE$.
A: HINT : 
We have $\angle{HAE}=\angle{HEA}$, so $\triangle{HAE}$ is an isosceles triangle with $HA=HE$. 
Now prove that $\triangle{HBE}$ is an isosceles triangle with $HB=HE$.
A: Let $\alpha = \frac12 \angle CAB$. We have $\alpha = \angle AEH$ (alternate interior angles), so $\triangle AEH$ is isosceles of vertex $H$, so $HA = HE$.
Let $\beta = \angle HBE$, $\gamma = \angle EBD$ and $\theta = \angle EDG$. We have:
$$\beta + \gamma = \pi- (\angle ACB + 2 \alpha) = \pi - \angle EGD - 2\alpha = \pi - (\pi - (\alpha + \theta)) - 2\alpha = \theta - \alpha = (\pi - \angle EDB)  - \alpha = \pi - (\pi/2 - \gamma) - \alpha = \pi/2 + \gamma - \alpha$$
Hence, $\beta = \pi/2 - \alpha = \angle HEB$, so $\triangle HEB$ is isosceles of vertex $H$ and $HE = HB$.
Therefore, $HB = HA$
A: We need to consider only the triangle $ABE$ with the point $H$ on $AB$ such that $\angle HEB=\angle EAC=\angle BAE=\theta$. We have $\angle BHE=2\theta$ and $\angle BEH=\frac12\pi-\theta$; so $\angle HBE=\frac12\pi-\theta$. Hence triangle $BHE$ is isosceles, as is $EHA$, with $|HB|=|HE|=|HA|.$
