Question

line segment $overline{cd}$ is the perpendicular bisector of line segment $overline{ab}$. which statement shows why $overline{ae}congoverline{be}$?
a $mangle efa = mangle efb=90^{circ}$ by definition of perpendicular so $angle efacongangle efb$
$overline{af}congoverline{bf}$ by the definition of midpoint
$\triangle eafcong\triangle ebf$ by hypotenuse - leg
$overline{ae}congoverline{be}$ because they are corresponding parts of congruent triangles
b $mangle efa + mangle efb = 180^{circ}$ so $\triangle aeb$ must be isosceles
$overline{ae}congoverline{be}$ because the legs of isosceles triangles are congruent
c $mangle efa = mangle efb = 90^{circ}$ by definition of perpendicular so $angle efacongangle efb$
$overline{af}congoverline{bf}$ by the definition of midpoint

Explanation:

Step1: Recall perpendicular - bisector properties

Since $\overline{CD}$ is the perpendicular bisector of $\overline{AB}$, by the definition of perpendicular, $m\angle EFA = m\angle EFB=90^{\circ}$, so $\angle EFA\cong\angle EFB$. Also, by the definition of mid - point, $\overline{AF}\cong\overline{BF}$. And $\overline{EF}\cong\overline{EF}$ by the reflexive property.

Step2: Prove triangle congruence

In right - triangles $\triangle EAF$ and $\triangle EBF$, we have $\overline{AF}\cong\overline{BF}$, $\angle EFA\cong\angle EFB$ and $\overline{EF}\cong\overline{EF}$. So, $\triangle EAF\cong\triangle EBF$ by the Hypotenuse - Leg (HL) congruence theorem for right - triangles.

Step3: Use corresponding parts of congruent triangles

Since $\triangle EAF\cong\triangle EBF$, corresponding parts of congruent triangles are congruent. So, $\overline{AE}\cong\overline{BE}$.

Answer:

A. $m\angle EFA = m\angle EFB = 90^{\circ}$ by definition of perpendicular so $\angle EFA\cong\angle EFB$, $\overline{AF}\cong\overline{BF}$ by the definition of mid - point, $\triangle EAF\cong\triangle EBF$ by Hypotenuse - Leg, $\overline{AE}\cong\overline{BE}$ because they are corresponding parts of congruent triangles.