Question

given: (overline{de}paralleloverline{gf},overline{ef}paralleloverline{dg}) prove: (mangle edg = mangle efg) 1. (overline{de}paralleloverline{gf},overline{ef}paralleloverline{dg}) 2. if two parallel lines are cut by a trans - versal, then their alternate interior angles are equal. 3. (mangle1 + mangle2=mangle4 + mangle3)

Explanation:

Step1: State the given

Given that $\overline{DE}\parallel\overline{GF}$ and $\overline{EF}\parallel\overline{DG}$

Step2: Identify equal angles

Since $\overline{DE}\parallel\overline{GF}$, $\angle 1$ and $\angle 4$ are alternate - interior angles, so $\angle 1=\angle 4$. Also, since $\overline{EF}\parallel\overline{DG}$, $\angle 2$ and $\angle 3$ are alternate - interior angles, so $\angle 2=\angle 3$.

Step3: Use angle - addition postulate

We know that $\angle EDG=\angle 1+\angle 2$ and $\angle EFG=\angle 3 + \angle 4$. Substituting $\angle 1=\angle 4$ and $\angle 2=\angle 3$ into the expressions for $\angle EDG$ and $\angle EFG$, we get $m\angle EDG=m\angle EFG$.

Answer:

1. Given
2. $\angle 1=\angle 4$, $\angle 2 = \angle 3$ (If two parallel lines are cut by a trans - versal, then their alternate interior angles are equal)
3. By the angle - addition postulate (If $\angle A=\angle C$ and $\angle B=\angle D$, then $\angle A+\angle B=\angle C+\angle D$)