Let $E$ be the center of side $CD$ of square $ABCD$. If the altitude in point $D$ on diagonal $BD$ intersects $AE$ in point $F$, prove that $B,C,F$.. 

Let $E$ be the center of side $CD$ of square $ABCD$. If the altitude in point $D$ on diagonal $BD$ intersects $AE$ in point $F$, prove that $B,C,F$ are colinear.

I think the best way to go about this is by proving $BF$ is parallel to $AD$, therefore proving that $B,C,F$ are colinear. I'm not sure how to go about proving this though, so any help is appreciated. (maybe if $FAD=AFB$)
 A: Just in case you are interested, here is a proof using vectors:
Consider $F$ to be the intersection of the lines $AE$ and $DF$ as described, but assuming nothing about the line $CF$.
Let $\overrightarrow{BA}=\underline{a}=\overrightarrow{CD}$ and $\overrightarrow{BC}=\underline{b}=\overrightarrow {AD}$.
Then we have $$\overrightarrow{BD}=\underline{a}+\underline{b}$$
$$\overrightarrow{DF}\parallel \overrightarrow{AC}\implies\overrightarrow{DF}=\lambda(\underline{b}-\underline{a})$$
Also, $$\overrightarrow{AE}=\underline{b}-\frac12\underline{a}\implies\overrightarrow{AF}=\mu(\underline{b}-\frac12\underline{a})$$
Then, $$\overrightarrow{BF}=\underline{a}+\mu(\underline{b}-\frac12\underline{a})=\underline{a}+\underline{b}+\lambda(\underline{b}-\underline{a})$$
Comparing coefficients of $\underline{a}$, we get $\lambda=\frac12\mu$.
Comparing coefficients of $\underline{b}$, we get $\mu=1+\lambda\implies\mu=2,\lambda=1$
Therefore $\overrightarrow{BF}=2\underline{b}$, and therefore $B, C, F$ are collinear.
Note that $ABCD$ didn't have to be a square - it could have been a rhombus.
