Chapter 7.1, Problem 14PSB

To determine

To prove : If right triangle is isosceles, then it must be a ${45}^0-{45}^0-{90}^0$ triangle.

Explanation of Solution

Given information : The following information has been given

  $\Delta ABC$ is a right triangle

  $\angle A={90}^0$

  $\angle B=\angle C$

Formula used : Sum of the angles of any triangle add up to ${180}^0$

Proof : In $\Delta ABC$ , we know that

  $\angle A={90}^0$

Now, we also know that the sum of the angles of any triangle add up to ${180}^0$

Thus, we know that

  $\angle A+\angle B+\angle C={180}^0$

Now, we substitute the value of $\angle A={90}^0$ in the equation given above. Thus, we get

  $\angle B+\angle C={90}^0$

Now, we know that $\angle B=\angle C$ . Thus, from these two equations, we can say that

  $\angle B=\angle C={45}^0$

Hence, it is a ${45}^0-{45}^0-{90}^0$ triangle