Answer
The proof is in the following form:
Column 1 of proof; Column 2 of proof.
Thus, we have:
1. Triangles DEF and MNP are similar, and DG and MQ are the altitudes; Given
2. Angle E is congruent to angle N; CASTC
3. Angle EGD and angle NQM are 90 degrees; perpendicular lines form right angles
4. Angle EGD is congruent to angle NQM; right angles are congruent
5. Triangle EDG is congruent to triangle NMQ; AA
6. $\frac{DG}{MQ} = \frac{DE}{MN}$; CSSTP
