Hello Arundev,
Consider the following diagram:
https://www.physicsforums.com/attachments/1763._xfImport
Without loss of generality, I have chosen a unit semicircle whose center is at the origin.
Point $P$ is $$(x,y)=\left(x,\sqrt{1-x^2} \right)$$.
The slope of line segment $A$ is:
$$m_A=\frac{\sqrt{1-x^2}-0}{x-(-1)}=\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}}$$
The slope of line segment $B$ is:
$$m_B=\frac{\sqrt{1-x^2}-0}{x-1}=-\frac{\sqrt{1-x^2}}{1-x}=-\sqrt{\frac{1+x}{1-x}}$$
As proven http://mathhelpboards.com/math-notes-49/perpendicular-lines-product-their-slopes-2953.html, two lines are perpendicular if the prodict of their slopes is $-1$.
$$m_Am_B=\left(\sqrt{\frac{1-x}{1+x}} \right)\left(-\sqrt{\frac{1+x}{1-x}} \right)=-1$$
Thus, we know line segments $A$ and $B$ are perpendicular, and so the triangle is a right triangle.
