ABC is Isosceles right triangle with $AB=BC$ $P$ , $Q$ are points on $AC$ such that $AP ^2 + CQ^2 = PQ^2$ What is the value of angle $PBQ (x) $? 
ABC is Isosceles right  triangle with $AB=BC$
           $P$  , $Q$ are points on  $AC$ such that 
  $$AP ^2  + CQ^2  =  PQ^2$$

What is the value  of angle $PBQ (x) $?

Thank you for help 
 A: Let $AC=1, AP=x, CQ=y.$ Then, $x^2+y^2 = (1-x-y)^2\Rightarrow  2x+2y-1=2xy$ or $y = \dfrac{3}{2x+2}-1=\dfrac{1-2x}{2-2x}.$ By theorem of Cosine, $BP^2 = \dfrac{1}{2}+x^2 - 2x\dfrac{1}{\sqrt{2}}\cos\dfrac{\pi}{4} = x^2-x+\dfrac{1}{2}$ and similarly $BQ^2 = y^2-y+\dfrac{1}{2}.$ Therefore, $$\cos\angle PQB = \dfrac{BP^2+BQ^2-PQ^2}{2BP\cdot BQ} = \dfrac{1-x-y}{2BP\cdot BQ} = \dfrac{2x^2-2x+1}{2-2x}\cdot \dfrac{1}{2\sqrt{(x^2-x+\frac{1}{2})\frac{2x^2-2x+1}{4(x-1)^2}}}=\dfrac{1}{\sqrt{2}},$$
So $\angle PBQ = \dfrac{\pi}{4}.$ 
A: Here is a geometrical approach. 
Construct a quarter circle centre $B$ with radius equal to $BA$ and $BC$. Choose any point $X$ on this quarter circle, and let $P$ be a point on $AC$ so that $AP=PX$ amd likewise let $Q$ be a point on $ AC$ so that $XQ=QC$.
Then by SSS, triangles $BPA$ and $BPX$ are congruent, and similarly so are triangles $BQC$ and $BQX$.
Therefore $\angle BXP=\angle BAP=45$ And $\angle BXQ=\angle BCQ=45$
Therefore $\angle PXQ=90\implies PQ^2=PX^2+QX^2$ by Pythagoras.
Hence the condition $AP^2+QC^2=PQ^2$ is satisfied by this construction.
Then it remains simply to observe that since $\angle ABP=\angle PBX$ and $\angle CBQ=\angle XBQ$ and that all four angles sum to $90$, then $\angle PBX+\angle PBQ=\angle PBQ=45$
A: Draw a half circle $\mathcal{C} $ with diameter $PQ$ outside of the triangle. Let circle with center at $P$ and radius $PA$ cuts $\mathcal{C} $ at $S$. Then 
$$QS^2 = QP^2- PS^2 = QS^2-AP^2 = CQ^2\Longrightarrow \triangle CQS {\rm \;\;is \;\;isoceles} $$
Easy calculation gives $\angle ASC = 135^{\circ}$ thus $S$ is on circle with center at $B$ and radius $BA$. Thus $BS = BA = BC$ so $\triangle CQB \cong \triangle SQB (sss)$ and $\triangle APB \cong \triangle SPB (sss)$. So 
\begin{eqnarray}
\angle QSP &=& \angle QBS +\angle PBS \\
&=& {1\over 2} \angle CBS + {1\over 2} \angle ABS \\
&=& {1\over 2} \angle CBA \\
&=& 45^{\circ}
\end{eqnarray}
A: Another way to do this is to observe that $P$ could be coincident with $A$. In which case $AP=0$, therefore $CQ^2=PQ^2$ thus $CQ=PQ$. That means $Q$ bisects $AC$ and that $BQ$ bisects the right angle, therefore $∠PBQ$ is $\pi/4$.
