acute-angled triangle

Question

I try to solve this problem: I have an acute triangle $ABC$ and its heights intersect at point $S.$ I drew the situation. I have to find the length of $AS$. I know that $\angle BAC=45^{\circ}$ and $|BC|=7.$ Have you got any ideas?

Thanks for your help!

Answer 1

$AP=PC$, $SP=PB$ and $\angle SPA=\angle BPC=90^{\circ}$ and thereafter $\triangle APS \cong \triangle CPB$. Thus $AS=BC=7$.