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

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

A: 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.
A: Synthetic proof:
Construction:
Drop perpendiculars MD and NE on to the side AB, and MF and NG on to the side AC. Please refer fig. Join SA.
It is easy to see that the trinagles BDM , MSN and NGC thus formed are all similar to the given triangle ABC. From data, AB = AC. Therefore all the triangles BDM , MSN and NGC are also isosceles right angled triangles with,
$BD = DM = p$
$MS = SN = q$
$NG = GC = r$
From Pythagoras theorem,we have 
$BM^2 = 2p^2$
$CN^2 = 2r^2$
$MN^2 = 2q^2$
But from data, we have
$BM^2 + CN^2 = MN^2$
$=>2p^2 + 2r^2 = 2q^2$
$=>p^2 + r^2 = q^2$ ------ 1
$MDAF$ and $NSFG$ are rectangles.------ from construction. (refer fig.)
Therefore,
$MD = AF = p$  and  $NG = SF = r ------- 2$
In right angled triangle AFS,
$AF^2 + SF^2 = AS^2$
  => AS = q ------------- From 1 and 2.
=> In triangle MAN , MS = SN = AS = q  which means S is the circumcentre of triangle MAN. Chord MN subtends a right angle at S ( MSN is similar to right angled triangle ABC). Therefore on the circumference , i.e, at A (the circumcircle of MAN passes through A!) it subtends 45 degrees.  QED
Please comment.
