Chapter 7.5, Problem 18PPS

To determine

To prove: Theorem 7.9 using paragraph proof.

  $\frac{TA}{GB}=\frac{RT}{EG}$ .

Explanation of Solution

Given:

  

  $\Delta RTS\sim \Delta EGF$

  $\overline{TA},\overline{GB}$ are angle bisectors.

Theorem 7.9 states that, length of corresponding angle bisectors of two similar triangles are in proportion with the corresponding sides of the similar triangles.

As per Theorem 7.9:

  $\frac{TA}{GB}=\frac{RT}{EG}$

  $\Delta RTS\sim \Delta EGF$

  $\Rightarrow \angle R\cong \angle E,\angle RTS\cong \angle EGF$

Half of angle will also be equal.

  $\Rightarrow \frac{1}{2}\angle RTS\cong \frac{1}{2}\angle EGF$

  $\Rightarrow \angle RTA\cong \angle EGB$

By $A-A$ similarity, $\Delta RTA\sim \Delta EGB$

By property of similar triangles, ratio of corresponding sides are equal.

  $\frac{TA}{GB}=\frac{RT}{EG}$