6. In $\triangle A B C, A B=A C=13$ and $B C=10 . P$ is a point on $B C$ with $B P<P C$, and $H, K$ are the orthocentres of $\triangle A P B$ and $\triangle A P C$ respectively. If $H K=2$, find $P C$.
(1 mark)
在 $\triangle A B C$ 中, $A B=A C=13$ 而 $B C=10 \circ P$ 是 $B C$ 上滿足 $B P<P C$ 的一點, 而 $H 、 K$ 分別是 $\triangle A P B$ 和 $\triangle A P C$ 的垂心。若 $H K=2$, 求 $P C$ 。
6. $\frac{37}{5}$
6. Note that $A H K$ is perpendicular to $B C, P H$ is perpendicular to $A C$ and $P K$ is perpendicular to $A B$. Hence it is easy to see that $\triangle P H K \sim \triangle A B C$. Let $M$ be the midpoint of $B C$ (which is also the midpoint of $H K)$. Then $P M=A M \times \frac{H K}{B C}=\sqrt{13^{2}-5^{2}} \times \frac{2}{10}=2.4$, and so $P C=P M+M C=2.4+5=7.4$.