Let EH represent the perpendicular line drawn from E to BC, i.e. the height. Now,
$\displaystyle \Delta EHC\sim \Delta OFC$
But, $\displaystyle OF=\frac{1}{2}EH$ (half the height).
Thus,
$\displaystyle \frac{OF}{EH}=\frac{1}{2}$.
Now use similar triangles,
$\displaystyle \frac{OF}{EH}=\frac{FC}{CH}$
Thus,
$\displaystyle \frac{FC}{CH}=\frac{1}{2}$
Thus,
$\displaystyle FC=\frac{1}{2}CH$
Thus,
$\displaystyle CH=2FC$
Thus,
$\displaystyle CH=2(14)=28$
Then,
$\displaystyle HF=CH-FC=14$
Then,
$\displaystyle BH+HF=26$
Thus,
$\displaystyle BH+14=26$
Thus,
$\displaystyle BH=12$
Finally Pythagorean theorem,
$\displaystyle BH^2+HE^2=BE^2$
Thus,
$\displaystyle (12)^2+HE^2=(37)^2$
Thus,
$\displaystyle 144+HE^2=1369$
Thus,
$\displaystyle HE^2=1225$
Thus,
$\displaystyle HE=\sqrt{1225}=35$