For a triangle $ABC$, prove that for a point $P$ on the angle bisector of $\angle BAC$,$|PL|=|PK|$ where $L,K$ are intersections of circles with sides In a triangle $ABC$, the point $P$ lies on the angle bisector of $\angle BAC$. We define a circle $\alpha$ with diameter $AP$, circle $\beta$ which passes through $B$ and is tangent to $\alpha$ at the point $E$, where it intersects $AB$, and circle $\gamma$ which passes through $C$ and is tangent to $\alpha$ at the point $F$, where it intersects $AC$. Then the points $K,L$ are defined as the intersections of $BC$ with $\beta$ and $\gamma$ respectively. The task is to prove that $|PL|=|PK|$.

In the case when $ABC$ is isosceles with $CB$ as the base, we can see that since $\alpha$ passes through $A$ and $P$ which lies on the angle bisector, it passes through $AC$ and $AB$ symmetrically, i.e. the distances from $A$ to $D$ (intersection of $\alpha$ and $AB$) and from $A$ to $E$ (intersection of $\alpha$ and $AC$) are the same. Then from the fact that the trinagle is isosceles, $\beta$ and $\gamma$ intersect $CB$ symmetrically, i.e. $|CL| = |KB|$. Hence, for $KLP$, the angle bisector of $\angle BAC$ is also the angle bisector of $\angle KPL$. $KLP$ is then similar to $ABC$ and therefore isosceles with $LK$ as base.
I have not been successful with generalising this proof to all triangles, however. It certainly cannot make use of symmetry in a similar manner. How should I approach this?
Thank you for your help.
 A: 
The point where circle $\alpha$ intersects $AB$ is the foot of the perpendicular from $P$ on $AB$, call this point $D$. Circle $\beta$ is tangent to $\alpha$ at point $D$ and thereafter the midpoint of $AP$, $D$, and the centre of $\beta$ are collinear. By similarity, the centre of $\beta$ lies on the line parallel to $AP$ through the point $B$. Extend $PD$ to meet this parallel line at point $Q$. $BQ$ is the diameter of circle $\beta$. Now, observe that $K$ is the foot of the perpendicular from $Q$ on $BC$.
So we reframe the problem as follows :
In $\triangle ABC$, the internal angle bisector of $\angle A$ is drawn and two parallel lines $l_{1}$ and $l_{2}$ to this line are drawn through $B$ and $C$ respectively. $P$ is any point on the angle bisector. Perpendiculars from $P$ on $AB$ and $AC$ intersect $AB$ and $AC$ at $D$ and $E$ respectively. When extended, $PD$ and $PE$ meet $l_{1}$ and $l_{2}$ at $Q$ and $R$ respectively. Perpendiculars $QK$ and $RL$ on $BC$ are drawn intersecting $BC$ at $K$ and $L$ respectively. Prove that, $PK=PL$.
Draw parallel lines to $AB$ and $AC$ through $P$ intersecting $BC$ at $M$ and $N$ respectively.
Now, $\frac {PM}{PN}=\frac {AB}{AC}$.
$PQ=AB\cdot\frac {PD}{AD}$ ($\triangle PDA\sim \triangle QDB$) and similarly $PR=AC\cdot\frac {PE}{AE}=AC\cdot\frac {PD}{AD}$.
$\Rightarrow \frac {PQ}{PR}=\frac {AB}{AC}=\frac {PM}{PN}$
$\angle MPQ=\angle NPR=90^{\circ}$ and hence $\triangle QPM\sim \triangle RPN$.
Quadrilaterals $KQPM$ and $NRPL$ are cyclic. So, $\angle PKM=\angle PQM=\angle PRL=\angle PNL$ and therefore $PK=PL$.
A: 
Hint: M is mid point of KL. You have to show trapezoids TPMK and MPNL are equal which results in PL=PK.
You may also use following figure:

In this figure M and N are circomcircles of triangles APK and APL respectively. AE is radical axis of these circles and EL=EB. If DE=KB then DK=EL and triangles PKD and OLE are equal that means PL=PK.
A: 
We reproduce the diagram given with the problem statement after extending it by adding the line segments $GB$, $GK$, $GD$, $HC$, $HL$, and $HD$, where $D$, $G$, and $H$ are the centers of the circles $\alpha$, $\beta$, and $\gamma$ respectively. We also draw the segments $KE$ and $LF$, and then extend both so that they meet at $N$. Please note that the line passing through $N$ and $P$, which intersects with $BC$ at $M$, is also added.
For brevity, let $\measuredangle CAB = 2\phi$. Furthermore, we have $\measuredangle BAP = \measuredangle PAC = \phi$, because $AP$ is the angle bisector of the angle $CAB$.
Let’s do some angle chasing. Since $DE$, $DA$, and $DF$ are radii of the circle $\alpha$, the triangles $EDA$ and $ADF$ are isosceles triangles. As a consequence,
$$\measuredangle AED = \measuredangle DFA = \phi. \tag{1}$$
Since, the angles $BEG$ and $HFC$ are vertically opposite to the angles $AED$ and $DFA$ respectively, we have from (1),
$$\measuredangle BEG = \measuredangle HFC = \phi. \tag{2}$$
Since $GB$ and $GE$ are radii of the circle $\beta$, the triangles $EGB$ is an isosceles triangle. Therefore, using (2), we shall write,
$$\measuredangle GBE = \measuredangle BEG = \phi. \tag{3}$$
Since $HC$ and $HF$ are radii of the circle $\gamma$, the triangles $CHF$ is an isosceles triangle. Therefore, as per (2), we get,
$$\measuredangle FCH = \measuredangle HFC =  \phi. \tag{4}$$
Now, pay attention to the circle $\beta$ exclusively. According to (3), the angle on the other side of the apex angle of the isosceles triangle $EGB$ can be expressed as
$$\measuredangle BGE = 180^o + \measuredangle GBE + \measuredangle BEG =180^o + 2\phi. \tag{5}$$
Angles $BKE$ and $BGE$ are subtended by the same arc of the circle $\beta$ at its circumference and the center respectively. From (5), it follows that
$$\measuredangle BKE = \frac{\measuredangle BGE}{2} = 90^o + \phi. \tag{6}$$
The angle $EKL$ is the supplement of the angle $BKE$. Hence, from (6), we have,
$$\measuredangle EKL = 90^o - \phi. \tag{7}$$
It is time to divert our attention to the circle $\gamma$. Using (4), the angle $CHF$ at the apex of the isosceles triangle $CHF$ can be written as
$$\measuredangle CHF = 180^o - \measuredangle FCH - \measuredangle HFC =180^o - 2\phi. \tag{8}$$
Angles $CLF$ and $CHF$ are subtended by the same arc of the circle $\gamma$ at its circumference and the center respectively. From (8), it follows that
$$\measuredangle CLF = \frac{\measuredangle CHF}{2} = 90^o - \phi. \tag{9}$$
According to (7) and (9), $\triangle LNK$ is an isosceles triangle. Its apex angle $LNK$ is given by
$$\measuredangle LNK = 180^o - \measuredangle NKL - \measuredangle KLN = 2\phi. \tag{10}$$
Therefore, we have $\measuredangle FNE = \measuredangle FAE$, which means that the point $N$ lies on the circle $\alpha$. Since $PNE$ and $PAE$ are angles in the same segment, $\measuredangle PNE = \measuredangle PAE = \phi$. Due to a similar reason, $\measuredangle FNP = \measuredangle FAP = \phi$ as well. Now, we have enough evidence, i.e. $\measuredangle PNE = \measuredangle FNP$, to state that $NM$ is the bisector of the angle $LNK$. Since $\triangle LNK$ is an isosceles triangle, $NM$ is perpendicular to $KL$, which makes the point $M$ the midpoint of the segment $KL$.
As the final step of this proof, consider the two right triangles $KMP$ and $PML$, where we have $\measuredangle KMP = \measuredangle PML = 90^o$. The side $MP$ is common to both triangles. Since $M$ is the midpoint of $KL$, we have $KM =ML$. According to SAS rule, the two triangles are congruent requiring $PK = PL$.
