$ABCD$ is a trapezoid with two right angles at $A$ and $B$. The diagonals intersect at $P$. The line through $P$ parallel to the bases intersects sides $AB$ and $CD$ at points $E$ and $F$, respectively.
Prove that $EF$ is the angle bisector of angle $\angle CED$.
If I can prove that the triangles $\Delta BEC$ and $\Delta AED$ are similar, I am done. just one angle apart from the right angle needs to shown to be equal. Any hints? Thanks beforehand.
On
Say $E =(0,0)$ and we can assume that $P=(1,0)$. Let the line $y=l(x-1)$ cuts y-axis at $A = (0,-l)$ and the line $y=k(x-1)$ cuts y-axis at $B=(0,-k)$. Then we have $$ C = (1-{k\over l},-k)\;\;\;{\rm and}\;\;\; D = (1-{l\over k},-l)$$
Now the slope of $EC$ is $$ k_1 = {-kl\over l-k}$$ and the slope of $ED$ is $$ k_2 = {-kl\over k-l} = -k_1$$ and we are done.
The triangles $\triangle(BCP)$ and $\triangle(DAP)$ are similar, hence their heights $EB$ and $EA$ satisfy $|EB|:|BC|=|EA|:|DA|$.