Finding and proving similar triangles ABC is a triangle with AB shorter than side AC. The angle bisector of ∠A intersect BC at D. Given that point E is on the median that's drawn from A, so that BE⊥AD, how do I show that DE||AB? I tried to prove similar triangles, but can't find a way to finish.
 A: Thanks for the interesting problem.  Here is a sketch of the proof.
Mark the middle point of $BC$ as $M$, so $E$ lies on $AM$.  Let the intersections of the line $BE$ with $AD$ and $AC$ be $H$ and $K$, respectively.  Finally, let us mark the middle point of $KC$ as $N$, and join $MN$. We shall show $DE \parallel AB$ by proving
$$
\frac{ME}{MA} = \frac{MD}{MB}.
\qquad (1)
$$

First, we can show that $MN \parallel BK$.  This is because in $\triangle CBK$,
$M$ and $N$ are the middle points of $CB$ and $CK$, respectively.
This means $EK \parallel MN$ in $\triangle AMN$, and it follows
$$
\frac{ME}{MA} = \frac{NK}{NA} = \frac{CN}{NA}.
\qquad (2)
$$
Second, we can similarly show that $HN \parallel BC$.  This is because in $\triangle KBC$, $H$ and $N$ are the middle points of $KB$ and $KC$, respectively.
It follows that, in $\triangle ADC$,
$$
\frac{CN}{NA} = \frac{DH}{HA}.
\qquad (3)
$$
Third, we can show that $HM \parallel CK$. Then in $\triangle DAC$,
$$
\frac{DH}{HA} = \frac{DM}{MC} = \frac{MD}{MB}.
\qquad (4)
$$
Combining (2), (3) and (4) yields (1).
