Find the measure of $\angle G$ in a triangle In the following figure, $\Delta$ADB, $\Delta$PCB and $\Delta$EFG are right triangles.
$PB=AE$,$AC=CB$.
Question:  What is the value of $\angle G$?
I figured out $\angle A = \angle CPB$ ,  $EF=AC$. What should I do next?

 A: We have similar right $\triangle PCB \cong \triangle AFE$ since $PB=AE$. Hence let $CB=x=EF=AC$. Also let $PC=y=AF$.

Then $CF=AF-AC=y-x$. Complete the rectangle $PCFQ$. We get $PQ=CF=y-x$. Also $QE=QF-EF=y-x$. Hence $\triangle PQE$ is right isosceles. Using similarity or since $PQ \parallel CG$, $$\angle G = 45^{\circ}$$
A: All you need is $AP = PB$ (as $AC = CB$) and so $AP = AE$.
If $\angle GAE = \theta, \angle PAE = 90^0 - 2\theta \implies  \angle APE = 45^0 + \theta$
That leads to $\angle CPG = \angle APE - \angle APC = 45^0$
So $\angle G = 45^0$
A: $CP$ is orthogonal bisector of segment $AB$. Hence
$$AP = BP = AE$$ which means that triangle $AEP$ is isosceles. Therefore
$$\angle \, AEP = \angle \, APE = \alpha \,\,\, \text{and} \,\,\, \angle \, PAE = 180^{\circ} - 2\alpha$$
Since $\angle \, DPE = 90^{\circ} - \angle \, AEP = 90^{\circ} - \alpha$,
$$\angle \, APB = \angle \, APE - \angle \, DPE = 2\alpha - 90^{\circ}$$
But $ABP$ is isosceles ($AP = BP$), so
$$\angle \, PAG = \angle  \, PAC = 90^{\circ} - \frac{1}{2}(\angle \, APB ) = 135^{\circ} - \alpha$$
From here,
$$\angle \, EAG = \angle \, PAG - \angle \, PAE = 135^{\circ} - \alpha - 180^{\circ} + 2\alpha = \alpha - 45^{\circ}$$
Now, in triangle $AEG$ we know the angle at vertex $A$, which is $\angle \, EAG = \alpha - 45^{\circ}$ and we know the external angle at vertex $E$, which is $\angle \, AEP = \alpha$. Theretore,
$$\angle \, G = \angle \, AGE = \angle \, AEP - \angle \, EAG = \alpha - \alpha + 45^{\circ} = 45^{\circ}$$
A: 
We can solve this problem without point $F$. As shown above, draw segment $PA$ and altitude $AH \perp PE$ , and let $AH$ meet $PC$ at $J$ .
Because $AC=CB \;\; \& \;\; PC \perp AB$ , triangle $PAB$ is isosceles and $PA = PB$ and $\widehat{APC} = \widehat{CPB}$
Points $P , D , C , A$ are on the same circle and as you already noted $\widehat{EAG} = \widehat{CPB}$ . Therefore we have
$$\widehat{APC} = \widehat{CPB} = \widehat{EAG} = \alpha$$
Because $AE = PB$ and $PB = PA$ , triangle $APE$ is isosceles and $\widehat{PAH} = \widehat{HAE}$.
Points $P , H , D , A$ are on the same circle and $\widehat{BPG} = \widehat{HAE}$ . Therefore we have
$$\widehat{PAH} = \widehat{HAE} = \widehat{BPG} = \beta$$
Note that
$$\widehat{G} = \widehat{PJH} = 90^o - \widehat{JPH} = 90^o - (\alpha + \beta)$$
Note also that $\widehat{PJH}$ is an external angle to $\triangle PJA$ :
$$\widehat{PJH} = \widehat{JPA} + \widehat{JAP} = \alpha + \beta $$
Therefore $\widehat{G} = \widehat{PJH} = 45^o$ .

We can also rephrase this problem as follows:
In isosceles triangle $PAB \;$ ($PA=PB$) we draw altitude $AD \perp PB$ and extend it to point $E$ such that $AE = PB$ . We then extend $PE$ and $AB$ to meet each other at point $G$.
Find $\widehat{G}$ .
