<s>[INST] In $Δ ABC,\;AB=AC$ and $\measuredangle CAB=90^o,\;M$ and $N$ are points on hypotenuse $BC$ with $BM^2+CN^2=MN^2$. Prove $\measuredangle MAN=45^o$. Let $ABC$ be a triangle in which $AB = AC$ and $|\angle CAB| = 90^{\circ}$. Suppose $M$ and $N$ are points on the hypotenuse $BC$ such that $BM^2 + CN^2 = MN^2$. Prove that $|\angle MAN| = 45^{\circ}$.

 [/INST] Let $\angle BAM=\alpha, \angle CAN=\beta, BM=x\sqrt{2},CN=y\sqrt{2}$.  We want to check that $\tan(\alpha+\beta)=1$.
The given formula is that $x^2+y^2=(1-x-y)^2$, or $1-2x-2y+2xy=0$.
$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\
=\frac{\frac x{1-x}+\frac y{1-y}}{1-\frac{xy}{(1-x)(1-y)}}\\
=\frac{x-xy+y-xy}{1-x-y}=\frac{1-x-y}{1-x-y}$$
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