Chapter 11, Problem 34RP

To determine

To prove: The area of triangle ABC is equal to area of triangle PQR .

Explanation of Solution

Given information:

  $\angle BAC$ is supplementary to $\angle QPR$ .

  $\Delta ABC$ and $\Delta PQR$ are isosceles with bases $\overline{BC}$ and $\overline{QR}$ .

Proof:

  

We can let $\angle BAC=2x$ , which will make $\angle QPR={180}^{\circ }-2x$ .

Thus $\angle DAC=x$ and $\angle SPR={90}^{\circ }-x$ .

They are thus complementary angles. But since $\Delta CDA$ is a right triangle, we know that $\angle C$ is complementary to $\angle DAC$ and thus $\angle C\cong \angle SPR$ .

  $\therefore \Delta ADC\cong \Delta RSP$ by AAS congruence rule and thus have equal areas.

Similarly, $\Delta BDA\cong \Delta PSQ$ and thus have equal areas.

  $\therefore A\left(\Delta ABC\right)=A\left(\Delta PQR\right)$