Chapter 7.5, Problem 12WE

To determine

To prove: For the given figure , $\frac{EF}{HI}=\frac{DF}{GI}$ .

Explanation of Solution

Given information:

  $\frac{DE}{GH}=\frac{EF}{HI};\angle E\cong \angle H$

  

According to SAS theorem;

Now consider one triangle has vertices D, E, and F and another triangle has vertices G, H, and I.

  $\begin{array}{l}\angle E\cong \angle H\\ \frac{DE}{GH}=\frac{EF}{HI}\end{array}$

Then, according to the SAS similarity theorem, if an angle of one triangle is congruent to an angle of another triangle and the sides including those angles are in proportion, then the triangles are similar.

Therefore,

  $\text{Δ}DEF-\text{Δ}GHI$

Now, as the triangles are similar, their corresponding sides must be in proportions,

  $\frac{DE}{DH}=\frac{DF}{GI}=\frac{EF}{HI}$

Hence, $\frac{EF}{HI}=\frac{DF}{GI}$