## Geometry: Common Core (15th Edition)

Yes, the polygons are similar. $ABDC ∼ DEFG$ The scale factor is $\frac{2}{3}$ or $2:3$.
First, we identify all the pairs of congruent angles: $\angle A ≅ \angle F ≅ \angle BDC ≅ \angle EDG$ $\angle B ≅ \angle E ≅ \angle C ≅ \angle G$ Now, let's take a look at the corresponding sides in both triangles: $\frac{AB}{DE} = \frac{4}{6}$ Divide the numerator and denominator by their greatest common factor, $2$: $\frac{AB}{DE} = \frac{2}{3}$ Let's look at $BD$ and $EF$: $\frac{BD}{EF} = \frac{4}{6}$ Divide the numerator and denominator by their greatest common factor, $2$: $\frac{BD}{EF} = \frac{2}{3}$ Let's look at $DC$ and $FG$: $\frac{DC}{FG} = \frac{4}{6}$ Divide the numerator and denominator by their greatest common factor, $2$: $\frac{DC}{FG} = \frac{2}{3}$ Let's look at $CA$ and $GD$: $\frac{CA}{GD} = \frac{4}{6}$ Divide the numerator and denominator by their greatest common factor, $2$: $\frac{CA}{GD} = \frac{2}{3}$ $ABDC ∼ DEFG$ because all angles are congruent, and all sides are proportional. The scale factor is $\frac{2}{3}$ or $2:3$.