2 circles intersect and points on it are taken such that the mid point of line joining them is an intersection point Let $\Gamma_1$ and $\Gamma_2$ be two circles with radius of $\Gamma_1$ smaller than that of $\Gamma_2$. Suppose they intersect at two distinct points A and B. Take a point C and $\Gamma_1$, and a point D on $\Gamma_2$
such that A is the midpoint of CD. Let CB be extended to meet circle $\Gamma_2$ at F, and DB be extended to meet $\Gamma_1$ at E. The perpendicular bisectors of CD and EF meet at P. Prove that
(a) $\angle EP F = 2 \angle CAE$; 
(b) $AP^2 = CA^2 + PE^2$

Let O be the centre of $\Gamma_1$ then we have to prove that triangle $COE$ is similar to triangle $EPF$. But how to do it? pls help.
 A: This answers the first question only.
Let (1) AP produced cut circle (BAD) at R and (2) CR cut (BAD) at T. Then, $\angle 1= \angle 2= \angle 3$. This means (CTD) has A as the center and CD the diameter with $\angle DTR = 90^0$.

The blue marked angles are equal.
The black dotted circle (EPF) cuts the perpendicular bisector of EF at S. Then, the red marked angles are equal. Also, the green marked angles are equal. Since (angle in red) + (angle in green) $= 90^0$, this makes $\angle PES = 90^0$.
Added info:------ Since PS is the diameter of (EPFS) and $\angle PAD = 90^0$ also, this means ADS is a straight line. $\angle PES = \angle PAS = 90^0$ implies A is also a con-cyclic point with the others.
Therefore, A is also a con-cyclic point on (EPFS) because $\angle PAD = 90^0$. This implies $\angle {brown} = \angle {red}$.  
Note that $\angle {blue} = 90^0 - \angle {brown} = 90^0 - \angle {red} = \angle {green}$. 
Required result of (1) follows.

A method in proving the said equality.
1) Produce PA to cut circle (CTD) at Q. 
2) Resolving AP and also PE into their corresponding pythagoras components, the target equality becomes $AS^2 + AQ^2 = QS^2$.
3) It remains to show S is the center of the circle passing through  E, F and also Q.
A: This is a late answer, no net a whole day, although there is already an accepted answer, a good one, i will still post the following, because of a constructive proof of (2), which i already arranged in geogebra. For (1) i have the same answer, but i will let it remain, same story a slightly different point of view.  

(1)
We have in the given situation:
$$
\begin{aligned}
\widehat{DAF}
&=
\widehat{DBF}
&&\text{ in the circle }DABF
\\
&=
\widehat{EBF}
\\
&=
\widehat{EAC}
&&\text{ w.r.t. the circle }AEBC
\ ,
\end{aligned}
$$
so the perpendicular in $A$ on the line $CAD$ is the angle bisector of the angle $\angle (EAF)$.
We draw the circumcircle $(EAF)$ of $\Delta EAF$, then the angle bisector $AP$ intersects the side bisector of $EF$ in the mid point of the arc 
$\overset \frown{EF}$. This is the point $P$. In particular, $AEPF$ cyclic. 
We construct the point $X$ on the circle $(AEPF)$, so that $PX$ diameter, i.e.
$\angle (PFX)=\angle (PEX)=\angle (PAX)=90^\circ$.
Because of $AP\perp CAD$ we  see that $X$ is also on the line $CAD$, 

We can then show the first needed relation,
$$
\begin{aligned}
\frac 12 \widehat{EPF}
&=
\widehat{XPF}
=
\widehat{XAF}
\\
&=
\widehat{DAF}
=
\widehat{DBF}
\\
&=
\widehat{EBF}
=
\widehat{CAE}
\ .
\end{aligned}
$$

(2)
I try now to show the equality
$$ PA^2 = PE^2 + AC^2$$
constructively, and in a manner that reveals as many properties of the objects in the figure as possible.
The strategy of proof can be followed in the picture: 



*

*(2.a) We show first $PE=PB$, but 
we need an intermediate object to negotiate better between $E,B$ for this,
it will be the point $Y$ constructed on the line $CBF$ so that $BED\|XY$.
We have 
$$
\widehat{DEF} =
\widehat{DBF} =
\widehat{XYF} \ ,
$$
so $Y$ is also on the circle $(AEPFX)$, a circle collecting an army of points. It follows that $YX$ is the angle bisector for $\angle(EYF)$, because $PX$ is the same for $\angle(EPF)$, and from $\angle(FYP)=\angle(FAP)$ we have three equal  "green angles" in $Y$. So $\Delta YEB$ isosceles. From $\Delta PEY=\Delta PBY$ we obtain $PE=PB$. 

*(2.b) We draw now the circle $(P)$ with center in $P$ and radius
$PE=PB$, it intersects $AP$ in two points, $T$ between $A$ and $P$,
and $T'\ne T$. This constructs $T,T'$. We finally construct the
point
$$
Z
$$
as the second intersection point ($\ne T'$) of $DT'$
with the circle $(P)=(TEBT')$. We are now in position to show...

*(2.c) The points $C,T,Z$ are colinear. To see this, we note that there is
an inversion with center $D$ that maps
$A\leftrightarrow C$,
$E\leftrightarrow B$,
$Z\leftrightarrow T'$. (The line $DEB$ through
the inversion center $D$ intermediates the equality of the powers
$DA\cdot DC=DE\cdot DB=DZ\cdot DT'$ between the two circles.)
In particular, $ACT'Z$ cyclic, so $CZ\perp ZT'$.
But also $TZ\perp ZT'$, because $TT'$ diameter in $(P)$. (It contains $P$.)
So $C,T,Z$ colinear.

*(2.d) $EF$ is a chord in the circle $(P)$, so $P$ is on its side bisector.

*(2.e) By construction $TZ\perp ZT'$, so $Z$ is not only
on the circle $(P)$, but also on the circle $(A)$ with
center $A$ and radius $AD=AC$.

*(2.f) The two circles $(A)$ and $(P)$ intersect orthogonally in $Z$,
$AZ\perp ZP$. For instance because of
$$
\begin{aligned}
2\widehat{AZP}
&=2(\pi-\widehat{AZD}-\widehat{PZT'})\\
&=2\pi -(\pi-\widehat{ZAD})-2\widehat{PZT'})\\
&=\pi +2\widehat{ZCD}-2\widehat{PZT'})\\
&=\pi \ .
\end{aligned}
$$
This implies (Pythagoras):
$$
\begin{aligned}
AP^2 &= AZ^2+ZP^2
\\
&=CA^2+PE^2\ ,
\end{aligned}
$$
the equality we need.
$\square$


*

*(2.g) Bonus: Let $W$ be the intersection $AP\cap EF$. Then $ZW\perp AP$,
i.e. $ZW\|DAC$.

