Chapter 8.3, Problem 9CE

To determine

To prove: $m\angle D=90$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem 8-2: The Square of the hypotenuse is equal to the sum of the squares of the base and perpendicular in a right triangle.

Theorem 8-3: If $c^2=a^2+b^2$ , then $m\angle C=90$ and $△ABC$ is right.

Consider the given figure:

Consider the right triangle $△ACB$ .

So, by theorem 8-2,

  $\begin{array}{c}A{B}^2=A{C}^2+C{B}^2\\ 3^2=n^2+2^2\\ 9=n^2+4\end{array}$

  $\begin{array}{c}9-4=n^2\\ 5=n^2\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{n^2}=\sqrt{5}\\ n=\sqrt{5}\end{array}$

Now, consider the triangle $△ADC$

Let the longest side of the triangle $△ADC$ be c, here $c=AC=\sqrt{5}$

Now,

  $\begin{array}{c}{c}^2={\left(\sqrt{5}\right)}^2\\ =5\end{array}$

Now find sum of the squares of other two sides $a=AD=\sqrt{3}$ , $b=DC=\sqrt{2}$ then

  $\begin{array}{c}{a}^2+b^2={\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2\\ =3+2\\ =5\end{array}$

It can be observed that,

  $5=5$

That is,

  $c^2=a^2+b^2$

So, by theorem 8-3, $△ADC$ is right.

And $m\angle D=90$