
Problem 8.2 Given an isosceles right angle triangle $A B C$ with $\Varangle A C B=90^{\circ}$. Point $P$ lies on $B C, M$ is the midpoint of $A B$ and let $L$ and $N$ be points from the segment $A P$ such that $C N \perp A P$ and $A L=C N$.

a) Find $\Varangle L M N$;

b) If the area of $\triangle A B C$ is 4 times greater than the area of $\triangle L M N$ find $\Varangle C A P$.


Solution: a) Let $\Varangle C A P=\alpha$. We find $\Varangle A C N=90^{\circ}-\alpha$ and $\Varangle M C N=\Varangle A C N-45^{\circ}=45^{\circ}-\alpha=\Varangle L A M$. Since $A M=C M$ and $A L=C N$ it follows that $\triangle A M L \cong \triangle C M N$. Therefore $\Varangle A M L=$ $\Varangle C M N$ and so $\Varangle L M N=90^{\circ}-\Varangle A M L+\Varangle C M N=90^{\circ}$.

b) It follows from $\triangle A M L \cong \triangle C M N$ that $L M=M N$ and since
$\Varangle L M N=90^{\circ}$ we get $S_{L M N}=\frac{M N^{2}}{2}$. Also $S_{A B C}=\frac{A C^{2}}{2}$. Now $\frac{4 M N^{2}}{2}=\frac{A C^{2}}{2}$ applies that $M N=\frac{A C}{2}$. Denote by $Q$ the midpoint of $A C$. Thus $Q M=Q N=\frac{A C}{2}=M N$. Therefore $\triangle Q M N$ is equilateral and $\Varangle Q N M=60^{\circ}$, which implies $\Varangle Q N A=60^{\circ}-45^{\circ}=$ $15^{\circ}$. But $A Q=Q N$ and so $\Varangle C A P=\Varangle Q N A=15^{\circ}$.
