Geometry question on right triangle Let ABC be a triangle in which $AB = AC$ and $\measuredangle CAB = 90^{\circ}$. Suppose $M$ and $N$ are points on the hypotenuse $BC$ such that $BM^2 + CN^2 = MN^2$. How do I  prove that $\measuredangle MAN = 45^{\circ}$.
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 $1-2x-2y+2xy=0$.
$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\
=\frac{\frac x{1-x}+\frac y{1-y}}{1-\frac{xy}{(1-x)(1-y)}}\\
=\frac{x-xy+y-xy}{1-x-y}=\frac{1-x-y}{1-x-y}$$
Without loss of generality, assume $BC=1$. Let $BM=x$ and $CN=y$, so $MN=1-x-y$. Then it is given that $x^2+y^2=(1-x-y)^2$. However, there is an infinite number of possibile pairs $(x,y)$ with $0&lt;x,y&lt;\frac{1}{2}$ such that $x^2+y^2=(1-x-y)^2$. Only 1 choice of $(x,y)$ leads to $\angle MAN=45$ degrees. In other words, you can not prove it.