Chapter 13.9, Problem 9WE

To determine

To prove: that an angle inscribed in a semicircle is a right angle.

Explanation of Solution

Given information:

A triangle inscribed in a circle of radius $a$ as shown below-

  

Calculation:

To show $\angle ACB$ is right angle it is sufficient to show both lines $\overline{AC}$ , and $\overline{BC}$ are perpendicular to each other implies equation of the circle passing through point $C=(b,c)$ . Now equation of the circle is $x^2+y^2=a^2$ , if it passes through $C=(b,c)$ , then it becomes $b^2+c^2=a^2$ .

Let us assume $\overline{AC},\overline{BC}$ are normal to each other, then,

  $\text{Slope of }\overline{AC}\times \text{Slope of }\overline{BC}=-1$

Now,

  $\begin{array}{l}\text{Slope of }\overline{AC}=\frac{c-0}{b+a}=\frac{c}{a+b}\\ \text{Slope of }\overline{BC}=\frac{c-0}{b-a}=\frac{c}{b-a}\end{array}$

And,

  $\begin{array}{l}\text{Slope of }\overline{AC}\times \text{Slope of }\overline{BC}=-1\\ \Rightarrow \frac{c}{a+b}\times \frac{c}{b-a}=-1\\ \Rightarrow c^2=-(b+a)(b-a)=-b^2+a^2\\ \Rightarrow b^2+c^2=a^2\end{array}$

Thus $C=(b,c)$ lies on the circle and $\angle ACB$ is a right angle.