Let △ A B C and E, D on [ A B ] and [...

Step 1
$\mathrm{\angle }ABD=\mathrm{\angle }EBD=\mathrm{\angle }ECD=\mathrm{\angle }ACE\Rightarrow$
$\mathrm{△}ABD\sim \mathrm{△}ACE\Rightarrow$
$\frac{AD}{AB}=\frac{AE}{AC}\Rightarrow$
$AD\cdot AC=AB\cdot AE$
$\mathrm{\angle }EPA=\mathrm{\angle }ECA=\mathrm{\angle }ECD=\mathrm{\angle }EBD=\mathrm{\angle }PBA\Rightarrow$
$\mathrm{△}EPA\sim \mathrm{△}PBA\Rightarrow$
$\frac{AE}{AP}=\frac{AP}{AB}\Rightarrow$
$A{P}^2=AB\cdot AE$
In the same way: $A{Q}^2=AC\cdot AD$
Then $A{P}^2=AB\cdot AE=AC\cdot AD=A{Q}^2$ , $AP=AQ$

Step 1
$\mathrm{\angle }AQR\ne \mathrm{\angle }ADP$ as mentioned in comment.I made two drawings, one isosceles. These points are noticeable:
The measure of the sides and angles of isosceles triangle APQ does not vary with the type of triangle.
It can be assumed as a kind of affine transformation. Because base of triangle is fixed and the location of vertex A alters. It is like moving a triangle APQ relative to the base of ABC to construct three circles with certain measures of diameter.