Question

given: $overline{ab}paralleloverline{fe}$; $overline{bc}$ bisects $angle abe$; $overline{de}$ bisects $angle bef$. prove: $mangle abc = mangle def$ 1. $overline{ab}paralleloverline{fe}$; $overline{bc}$ bisects $angle abe$; $overline{de}$ bisects $angle bef$. 1. given 2. $mangle abc=\frac{1}{2}mangle abe$ and $mangle def=\frac{1}{2}mangle bef$ 2. 空缺 3. $mangle abe = mangle bef$ 3. 空缺

Explanation:

Step1: State given information

Given $\overline{AB}\parallel\overline{FE}$, $\overline{BC}$ bisects $\angle ABE$, $\overline{DE}$ bisects $\angle BEF$.

Step2: Apply angle - bisector definition

$m\angle ABC=\frac{1}{2}m\angle ABE$ and $m\angle DEF=\frac{1}{2}m\angle BEF$ by angle - bisector definition.

Step3: Use parallel - line property

Since $\overline{AB}\parallel\overline{FE}$, $m\angle ABE = m\angle BEF$ (alternate interior angles).

Step4: Substitute and conclude

Substitute $m\angle ABE$ for $m\angle BEF$ in angle - bisector equations to get $m\angle ABC = m\angle DEF$.

Answer:

1. Given: The reasons for the first - step statements are given in the problem itself.
2. Definition of angle - bisector: By the definition of an angle - bisector, if a ray bisects an angle, it divides the angle into two equal - measure angles. So, if $\overline{BC}$ bisects $\angle ABE$, then $m\angle ABC=\frac{1}{2}m\angle ABE$, and if $\overline{DE}$ bisects $\angle BEF$, then $m\angle DEF = \frac{1}{2}m\angle BEF$.
3. Alternate interior angles: Since $\overline{AB}\parallel\overline{FE}$, by the property of alternate interior angles (when two parallel lines are cut by a transversal, the alternate interior angles are congruent), $\angle ABE$ and $\angle BEF$ are alternate interior angles formed by the parallel lines $\overline{AB}$ and $\overline{FE}$ with transversal $\overline{BE}$, so $m\angle ABE=m\angle BEF$.
4. Substitution: Substitute $m\angle ABE=m\angle BEF$ into the equations $m\angle ABC=\frac{1}{2}m\angle ABE$ and $m\angle DEF=\frac{1}{2}m\angle BEF$. We get $m\angle ABC = m\angle DEF$.