Compare the side lengths of the triangles by using Theorem 12-3.
See solution.
Practice makes perfect
We want to show that △ GDC ~ △ GFE, given that ⊙ A and ⊙ B have common tangents DF and CE.
As we can see, the tangent segments to ⊙ A and ⊙ B share a common endpoint outside the circles. By Theorem 12-3, if two tangent segments to a circle share a common endpoint outside the circle, then the two segments are congruent.
Therefore, GC is congruent to GD as well as GE is congruent to GF.
GC ≅ GD and GE ≅ GF
Recall that congruent segments have the same length. As a consequence, by the Division Property of Equality, the ratio of GC to GD and the ratio of GE to GF will be 1.
GC/GD=1 and GE/GF=1
By the Transitive Property of Equality, this shows that GCGD is equal to GEGF.
GC/GD=GE/GF
Now that we have a proportion, we can consider the Properties of Proportions. By switching the means, we can write an equivalent proportion.
GC/GD=GE/GF ⇒ GC/GE=GD/GF
Next, if we look at ∠ DGC and ∠ FGE, we can see that they are vertical angles.
Therefore, by the Vertical Angles Theorem, ∠ DGC and ∠ FGE are congruent angles.
∠ DGC ≅ ∠ FGE
Combining all this information, we can conclude that an angle of △ GDC is congruent to an angle of △ GFE and the sides that include the two angles are proportional. Finally, by the Side-Angle-Side (SAS) Similarity Theorem, △ GDC is similar to △ GFE.
△ GDC~ △ GFE
Let's summarize the above process in a flow proof.
Completed Proof
2
&Given:&& ⊙ A and ⊙ B with common
& &&tangents DF and CE
&Prove:&& △ GDC ~ △ GFE
Proof:
