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 $\measuredangle MAN=45^o$.

Question

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 $|\angle MAN| = 45^{\circ}$.

Answer 1

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 $0<x,y<\frac{1}{2}$ 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.

Answer 2

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 $1-2x-2y+2xy=0$. $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\ =\frac{\frac x{1-x}+\frac y{1-y}}{1-\frac{xy}{(1-x)(1-y)}}\\ =\frac{x-xy+y-xy}{1-x-y}=\frac{1-x-y}{1-x-y}$$

Answer 3

As sides AB=AC and angle BAC is 90 the triangle is a right angle isosceles triangle.

as $$BM^2+CN^2=MN^2$$ The side BM and CN can be folded to outside to join.image

The sides will join and give a right triangle. the line MA and AN will give a polygon.With the help of this you may prove that the Angle MAN is 45. This condition is only possible if BM=NC.

Take points close to each other you will find it is not possible

Answer 4

Assume $$∠MAN ≠ 45° \tag1$$ Let $CN’ = 0, BM’ = BC/2$, so $$CN’^2 + BM’^2 = M’N’^2 \tag2$$

Since $∠CAB = 90°, BM’ = CM’ = N’M’$, so $∠M’AN’ = 45°$.

Hence $(1)$ cannot be true if condition $(2)$ is to be met (Reductio ad absurdum).