Find the angle x in the figure below For reference: In the figure , P and T are points of tangency . Calculate x. (Answer:$90^o$}

Does anyone have any ideas? I couldn't find a way
$PODT$ is cyclic
$\angle PDO \cong \angle DOT\\
\triangle POA :isosceles\\
\triangle OTB: isosceles$
$DO$ is angle bissector $\angle B$
DT = DP

 A: Extend $AP$ and $BT$ to $K$. Then notice that $E$ is an orthocenter of triangle $ABK$ so we only need to prove $D\in KE$.

*

*Easy angle chase we see that $\angle PDT = 2\angle PKT$ and since $PD = TD$ we see that $D$ is a circum centre for $PETK$ so $D$ halves $KE$ and thus it lies on $KE$.


A: In this answer I assume that the radius of the circle is 1 and the origin is the centre of the circle.
Drawing tangents from the radiuses at $\theta$ and $\phi$ we get two lines ($DP$ and $DT$ in the diagram)
$$
\begin{align}
x\cos\theta+y\sin\theta&=1\\
x\cos\phi+y\sin\phi&=1
\end{align}
$$
Solving for the $x$ coordinate of the intersection of these two lines, $D$, gives
$$
x={\sin\theta-\sin\phi\over \sin(\theta-\phi)}.
\tag{1}
$$
The line from $(1,0)$ to the radius of $\theta$ ($PB$ in the diagram) has equation
$$
(x-1)\cos(90-\theta/2)+y\sin(90-\theta/2)=0
$$
and similarly the line from $(-1,0)$ to the radius of $\phi$ ($AT$ in the diagram) is
$$
(x+1)\cos(\phi/2)-y\sin(\phi/2)=0
$$
Solving for the $x$ coordinate of the intersection of these two lines, $E$, gives
$$
\begin{align}
x&={\cos(\theta/2+\phi/2)\over\cos(\theta/2-\phi/2)}\\
&=
{2\sin(\theta/2-\phi/2)\cos(\theta/2+\phi/2)
\over
2\sin(\theta/2-\phi/2)\cos(\theta/2-\phi/2)}
\\
&=
{
\sin\theta-\sin\phi
\over
\sin(\theta-\phi)
}
\end{align}
$$
where I have used the relation $$\sin((a+b)/2)\cos((a-b)/2)=\sin a+\sin b$$
in the numerator.
This is the same $x$ value as in (1), the point of intersection $D$, so the line $ED$ is parallel to the $y$ axis, and thus the angle of the line $ED$ with $AB$, the $x$ axis, is $90^\circ$.
A: Extend DC and mark D' symmetrical D. M = $ DT \cap AB$. The triangle DMD' is isosceles and as DM = D'M, MC will be the height of the triangle DMD' and so we will have x = 90 degrees

