Consider this configuration in the complex plane, and $(O)$ is the unit circle Let $z=e^{i\theta}$, then $x=yz$ We have $$t=\frac{2xy}{x+y}=\frac{2x}{z+1}$$Since a dilation at $T$ with scale factor $r$ sends $A$ to $A'$, we have $$\frac{a'-t}{a-t}=r$$or $$a'=ar+t(1-r)=ar+\frac{2x}{z+1}(1-r)$$$$\overline{a'}=\overline{a}r+\frac{2z}{x(z+1)}(1-r)$$Since $PX \perp OX$, $A'P \parallel OX$, we have $$\frac{p-x}{\overline{p}-\overline{x}}=-x^2, \frac{p-a'}{\overline{p}-\overline{a'}}=x^2$$Solve the above system give us $$p=\frac{2x+a'-x^{2}\overline{a'}}{2}=\frac{-\overline{a}rx^{2}+(\frac{4+2zr-2r}{z+1})x+ar}{2}$$Suppose that two points $P$ is the same for points $X_1$ and $X_2$ By Vieta theorem, we have $$x_{1}+x_{2}=\frac{4+2zr-2r}{\overline{a}r(z+1)}$$Therefore $$x_{1}x_{2}=\frac{x_{1}+x_{2}}{\overline{x_{1}+x_{2}}}=\frac{4+2zr-2r}{\overline{a}r(z+1)} \div \frac{4z+2r-2zr}{ar(z+1)}=\frac{a(2+zr-r)}{\overline{a}(2+r-zr)}$$We now have $$p=\frac{\overline{a}r\left(\frac{a(2+zr-r)}{\overline{a}(2+r-zr)}\right)+ar}{2}=\frac{ar(z+1)}{2z+r-zr}$$Since $$\frac{p-a}{p} \div \frac{x-t}{x-y}=\frac{2r-2}{r} \in \mathbb{R}$$we have $\measuredangle TXY = \measuredangle APO$