15. (2 marks) In $\triangle A B C, D, E$ and $F$ are respectively the midpoints of $A B, B C$ and $C A$. Furthermore $A B=10, C D=9, C D \perp A E$. Find $B F$.
(2 分) 在 $\triangle A B C$ 中, $D, E$, 和 $F$ 分別為 $A B, B C$ 和 $C A$ 的中點。且 $A B=10, C D=9, C D \perp A E$ 。求 $B F$ 。
15. As shown in the figure, $A B=10, B C=2 x$ and $C A=2 y . D, E$ and $F$ are midpoints of $A B$, $B C$ and $C A$ respectively, $\mathrm{G}$ is the centroid, and further $C D \perp A E$, we have $C D=9$, from the property of centroid $\frac{C G}{G D}=\frac{A G}{G E}=\frac{B G}{G F}=2$, hence $C G=6, G D=3$. Let $G E=a$, then $A G=2 a$. By Pythagorean theorem, we have $a^{2}+36=x^{2},(2 a)^{2}+36=(2 y)^{2}, 9+(2 a)^{2}=25$. If follows that $a=2, x=2 \sqrt{10}$ and $y=\sqrt{13}$. Let $\theta=\angle B C A$. Apply cosine rule to $\triangle B C A$, get $100=(2 y)^{2}+(2 x)^{2}-2(2 x)(2 y) \cos \theta$, so $x y \cos \theta=14$. Apply cosine rule to $\triangle B C F$, get $B F^{2}=y^{2}+(2 x)^{2}-2 y(2 x) \cos \theta=13+160-4 \times 14=117$. Hence $B F=\sqrt{117}=3 \sqrt{13}$.

Alternate Solution

Choose the coordinates of $A, B$ and $C$ as $(0,0),(10,0)$ and $(b, c)$ respectively. Then the coordinates of $D, E$ and $F$ are $(5,0),\left(\frac{10+b}{2}, \frac{c}{2}\right)$ and $\left(\frac{b}{2}, \frac{c}{2}\right)$ respectively.

As $C D=9$, hence $(b-5)^{2}+c^{2}=81$, or $b^{2}+c^{2}=10 b+56$.
Also $C D \perp A E$, get $\frac{c}{10+b} \cdot \frac{c}{b-5}=-1$, or $b^{2}+c^{2}=-5 b+50$.
(1) and (2) imply $b=-\frac{2}{5}$.

Finally $B F^{2}=\left(\frac{c}{2}\right)^{2}+\left(\frac{b}{2}-10\right)^{2}=\frac{1}{4} c^{2}+\frac{1}{4}(b-20)^{2}$
$$
=\frac{1}{4}\left(b^{2}+c^{2}\right)-10 b+100=\frac{1}{4}(52)+4+100=117 .
$$

Hence $B F=\sqrt{117}=\underline{\underline{3 \sqrt{13}}}$