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I have posted a link there to this thread so the OP can view my work.Using coordinate geometry prove that angle in a semicircle is a right angle?

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I have posted a link there to this thread so the OP can view my work.Using coordinate geometry prove that angle in a semicircle is a right angle?

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Consider the following diagram:

Without loss of generality, I have chosen a unit semicircle whose center is at the origin.

Point $P$ is \(\displaystyle (x,y)=\left(x,\sqrt{1-x^2} \right)\).

The slope of line segment $A$ is:

\(\displaystyle m_A=\frac{\sqrt{1-x^2}-0}{x-(-1)}=\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}}\)

The slope of line segment $B$ is:

\(\displaystyle m_B=\frac{\sqrt{1-x^2}-0}{x-1}=-\frac{\sqrt{1-x^2}}{1-x}=-\sqrt{\frac{1+x}{1-x}}\)

As proven >>>here<<<, two lines are perpendicular if the prodict of their slopes is $-1$.

\(\displaystyle m_Am_B=\left(\sqrt{\frac{1-x}{1+x}} \right)\left(-\sqrt{\frac{1+x}{1-x}} \right)=-1\)

Thus, we know line segments $A$ and $B$ are perpendicular, and so the triangle is a right triangle.