Chapter 9.2, Problem 50WE

To determine

To prove: An angle inscribed in a semicircle is a right angle.

Explanation of Solution

Given information: The semicircle is the top half of the circle $x^2+h^2=r^2$ and $P\left(x,y\right)$ is any point on the semicircle. And the given coordinates are $A\left(-r,0\right)\&B\left(r,0\right)$ .

Formula used: $mXY=\frac{y_2-y_1}{x_2-x_1}$

Proof: Calculate the slope of line PA and PB,

  $\begin{array}{l}mPA=\frac{y-0}{x+r}=\frac{y}{x+r}\\ mPB=\frac{y}{x-r}\end{array}$

Now, get the value of $x+r$ ,

  $\begin{array}{l}{x}^2+y^2=r^2\\ x^2-r^2=-y^2\\ \left(x+r\right)\left(x-r\right)=-y^2\\ x+r=\frac{-y^2}{x-r}\end{array}$

Hence, $mPA$ :

  $\begin{array}{l}mPA=\frac{y}{\frac{-y^2}{x-r}}\\ =-\frac{x-r}{y}\end{array}$

As we know, the product of slopes of two segments that are perpendicular is $-1$ .

  $\begin{array}{l}mPA\times mPB=-\frac{x-r}{y}\times \frac{y}{x-r}\\ mPA\times mPB=-1\end{array}$

Therefore, the given segments are perpendicular to each other which proves that an angle inscribed in a semicircle is a right angle.