Chapter 8.3, Problem 21WE

To determine

To prove: that $\Delta ABC$ is a right triangle.

Explanation of Solution

Given information:

The given figure is as follows:

  

  $CN\perp AB$

  $h$is the geometric mean between $d$ and $e$ .

Concept used:

Pythagorean theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of its base and altitude.

Proof:

  $\Delta CNB$ is a right angled triangle.

Applying Pythagorean theorem, you get

  $C{B}^2=h^2+e^2$

Also, $\Delta CNA$ is a right angled triangle.

Applying Pythagorean theorem, you get

  $A{C}^2=h^2+d^2$

Given, $h$is the geometric mean between $d$ and $e$ .

So, $h^2=ed$

Now, consider $A{C}^2+C{B}^2$ ,

Substituting the obtained values,

  $\begin{array}{l}A{C}^2+C{B}^2\\ =h^2+d^2+h^2+e^2\\ =2h^2+d^2+e^2\\ =2de+d^2+e^2\\ $ ={(e+d)}^2$=A{B}^2\\ \Rightarrow A{C}^2+C{B}^2=A{B}^2\end{array}$

That is, $\Delta ABC$ satisfies the Pythagorean theorem.

Therefore, $\Delta ABC$ is a right triangle.

Hence, proved.