Apollonius circles theorem proof 
Given the base of a triangle and ratio of other $2$ sides. Find the locus of the third vertex

To prove that a geometric shape is the correct locus for a given set of conditions. We have to divide the proof into two stages:


*

*Proof that all the points that satisfy the given conditions are on the given shape.

*Proof that all the points on the given shape satisfy the given conditions. 


I am able to prove that the locus of a point which satisfy the satisfy the given conditions is a circle.
Let $BC$ be the base. And $A$ be the third vertex. We are given $AB:AC$ ratio. Angle Bisectors of $\angle A$ meet $BC$ at $P$ and $Q$. Since $\angle PAQ$ is a right angle. The locus of $A$ is a circle with $PQ$ as a diameter. This is first proof.
I couldn't obtain the solution for second proof. I want to prove that all the points on a circle with $PQ$ as a diameter is such that the ratio of the other two sides is constant that we initialised earlier. We initialised $AB:AC$ to be constant. Let a new point on the circle be $A'$. I want to prove that $A'B:A'C$ is same as $AB:AC$.
 A: At this moment, I can only offer the following particular solution to your problem. 
In $\triangle ABC$, we have $\theta = \theta’$ such that, by angle bisector theorem, AB : AC = BP : PC.
The black circle with PQ as diameter is constructed as described. 

Let AC be extended to cut the black circle at A’ (which will be our particular point on the locus). Then, $\theta = \theta_1$.
Next A’P is extended to cut QA extended at R.
Then, $\angle 1 = \angle 2 = \angle 3 = \angle 4$ implies APBR is cyclic. Then, x = y and $\theta’ = \theta’_1$.
$\theta'_1 = ... = \theta_1$ further means BRQA’ is cyclic with x = z.
z = y means P is the in-center of $\triangle ABA’$. Result follows.
Of course, if we use $A_1$, the diametric replica of A, as the third vertex, we will generate the second instance, namely A’’.
For the future development, I think the line BR will be of great help because $BR \bot BQ$.

Added.

Let X be a point on the said locus (i.e. the black circle with PQ as a diameter). We further let X’ be the reflection of X about PQ. As a consequence of BQ being the perpendicular bisector of XX’, we have (1) the purple marked angles are equal; and (2) $\rho_1 = \rho_2$. 
Form the rays XP and XC. Suppose that XC extended cuts BX’ at T. [You need to show that T is also a con-cyclic point of the black circle. Geogebra confirms that is true.]
Then, $\rho_3 = \rho_1 = \rho_2 = \rho_4$. This means p is the in-center of $\triangle XBT$. Result follows.
A: Let points $P$ and $Q$ on line $BC$ satisfy $PB/PC=QB/QC$ and construct the circle of diameter $PQ$ (red circle in diagram below). Draw now a circle of center $B$ and radius $1$ (dashed in the diagram) and consider the inversion transformation with respect to that circle. Points $P$ and $Q$ are transformed into $P'$ and $Q'$, the red circle of diameter $PQ$ into the blue circle of diameter $P'Q'$, and point $C$ is transformed into $C'$, midpoint of $P'Q'$ (this follows from the above relation among points $BCPQ$).
A point $A$ on the red circle is transformed into point $A'$ on the blue circle: from $BA\cdot BA'=BC\cdot BC'$ it follows $BA':BC'=BC:BA$, and triangles $BAC$ and $BC'A'$, having an angle in common, are similar. We then get $AC:A'C'=AB:BC'$, whence:
$$
{AB\over AC}={BC'\over A'C'}={BC'\over P'C'}=
{BC'\over BP'-BC'}={1/BC\over 1/BP-1/BC}=
{BP\over BC-BP}={BP\over CP},
$$
which is what we wanted to prove.

A: Here's another way to get the same result. Let $M$ be the midpoint of $PQ$ (and center of the circle), $\theta=\angle AMB$ and $r=PM=QM=AM$. By the cosine rule applied to triangles $ABM$ and $ACM$ we have:
$$
AB^2=BM^2+r^2-2r\, BM\,\cos\theta,\\
AC^2=CM^2+r^2-2r\, CM\,\cos\theta.
$$
Substituting here $BM=r+BP$ and $CM=r-CP$, we get after some algebra:
$$
AB^2={1\over2}(1+\cos\theta)BP^2+{1\over2}(1-\cos\theta)(2r+BP)^2,\\
AC^2={1\over2}(1+\cos\theta)CP^2+{1\over2}(1-\cos\theta)(2r-CP)^2.
$$
But $2r+BP=QB$ and $2r-CP=QC$,
thus we may rewrite the above equalities as follows:
$$
AB^2={1\over2}BP^2\left[1+\cos\theta+
(1-\cos\theta)\,\left({QB\over BP}\right)^2\right],\\
AC^2={1\over2}CP^2\left[1+\cos\theta+
(1-\cos\theta)\,\left({QC\over CP}\right)^2\right].
$$
As we know by hypothesis that $PB:PC=QB:QC$, the two expressions in square brackets are equal between them, and we obtain: 
$$AB:AC=BP:CP,$$ 
which is what we wanted to prove.

A: Let’s assume that $A'$ is a point on the Apollonius circle of triangle $\triangle ABC$. In order to show that $\frac{A' B}{A' C}$ is equal to the constant ratio $k=\frac{AB}{AC}$, let's choose a point $C'$ on BQ such that $A'P$ is the bisector of the $\angle BA'C'$. Now if $C=C'$ we are done and the proof is complete. If not then assume  $C'$ is on the right side of the point $C$. Because $A'$ is on the circumcircle of $\triangle PA'Q$ therefore $\angle PA'Q=90^{\circ} $ so $A'Q$ is the external bisector of $\angle BA'C'$. Now we can write the bisector related proportions:
In $\triangle ABC$:    $\;\;\;\;\frac{PB}{PC}=\frac{QB}{QC} \;\;\;\;\;\;\;\;\;\;\;\;$
And in $\triangle A'BC'$:    $\;\ \frac{PB}{PC'}=\frac{QB}{QC'}$
By dividing the proportions we get $$\frac{PC}{PC'}=\frac{QC}{QC'}$$
The left side is less than one and the right side is greater than 1, which is a contradiction. With a similar argument $C'$ cannot be on the left side of the $C$, therefore $C=C'$ hence $\frac{A'B}{A'C}=\frac{AB}{AC}=k$
