Two prove two lines in a triangle are parallel

Question

$D$, $E$, $F$ are the midpoints of sides $BC$, $CA$ and $AB$ respectively of a triangle $ABC$ right angled at $C$. If $EF$ and $DF$ (extended if necessary), meet the perpendicular from $C$ on $AB$ in points $G$ and $H$ respectively, show that $AG$ is parallel to $BH$.

Answer 1

After fixing the typo, the correct figure is:-

By midpoint theorem, 1) $GE // BC$; 2) $GE$ and $HD$ are the perpendicular bisectors of $AC$ and $BC$ respectively.

$\epsilon + \theta = 2 \theta = 2 \mu = \mu + \rho = \phi$

Result follows.

Answer 2

This is not a solution yet. It is just a picture to clarify a few things.

Answer 3

construction: extend CE to meet BH in J and extend AG to meet CE in I

Now consider the triangle ACE.We have CK is perpendicular to AE and EF is perpendicular to AC.Therefore G is the orthocentre of triangle ACE.Therefore, AI is another altitude.So we have angle AIE is a right angle.

Now consider the triangle BHC.By a similar argument we can prove that E is orthocentre of triangle BHC. Therefore, angle CJH is a right angle.

Thus,it is established that angle AIE=angle HJE.Therefore, AG is parallel to BH.