Chapter 8.4, Problem 1PSA

To determine

To prove : The equality of $\frac{AB}{BC}$ with respect to $\frac{DE}{EF}$ .

Explanation of Solution

Given information : Here, we have given that $\angle C\cong \angle F$ , $\overline{AB}\perp \overline{BC}$ and $\overline{DE}\perp \overline{EF}$ .

Proof: As, $\overline{AB}\perp \overline{BC}$ ,

  $\Rightarrow \angle ABC=90°$

Similarly, as $\overline{DE}\perp \overline{EF}$ ,

  $\Rightarrow \angle DEF=90°$

  $\Rightarrow \angle ABC\cong \angle DEF$

Now, In $\Delta ABC$ and $\Delta DEF$

  $\angle C\cong \angle F$ (Given)

  $\angle ABC\cong \angle DEF$ (Proved above)

  $\Rightarrow \Delta ABC\sim \Delta DEF$ (By AA Similarity Rule)

  $\Rightarrow \frac{AB}{DE}=\frac{BC}{EF}$ (Corresponding sides of similar triangles are equal in ratio)

Or $\frac{AB}{BC}=\frac{DE}{EF}$