Chapter 8.3, Problem 17PSB

To determine

To prove : The similarity of the triangles formed when perpendiculars are drawn to the base.

Explanation of Solution

Given information : Here, we have given that, $\overline{DF}\perp \overline{BC}$ and $\overline{EG}\perp \overline{BC}$

  

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

  $\Rightarrow \angle DFB=90°$

Similarly, as $\overline{EG}\perp \overline{BC}$ ,

  $\Rightarrow \angle EGC=90°$

  $\Rightarrow \angle DFB\cong \angle EGC$

Now, as $\Delta ABC$ is an isosceles triangle,

  $\Rightarrow \overline{AB}\cong \overline{AC}$

  $\Rightarrow \angle ABC\cong \angle ACB$ (Angles opposite to equal sides are equal)

Now, In $\Delta DFB$ and $\Delta EGC$

  $\angle DFB\cong \angle EGC$ (Proved above)

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

  $\Rightarrow \Delta DFB\sim \Delta EGC$ (By AA Similarity Rule)

  $\therefore$ Triangles formed are similar