Chapter 4.7, Problem 29WE

To determine

To prove: that $\overline{BE}\cong \overline{CF}$ with the help of given information.

Answer to Problem 29WE

  $\overline{BE}\cong \overline{CF}$ by Angle-Side-Side postulate.

Explanation of Solution

Given information:

Given: $\overrightarrow{AM}$ is perpendicular bisector of $BC$

  $\overline{AE}\perp \overline{BD}$

  $\overline{AF}\perp \overline{DF}$

  $\angle 1\cong \angle 2$

Calculation:

Consider the following data:

Prove that: $\overline{BE}\cong \overline{CF}$

In $△AED$ and $\text{Δ}AFD$

  $\angle AED\cong \angle AFD\overrightarrow{AE}\perp \overrightarrow{BD};\overrightarrow{AF}\perp \overrightarrow{DF}$

Given $\angle 1\cong \angle 2$

  $\overrightarrow{AD}\cong \overrightarrow{AD}$ Reflexive Property

Therefore, by Angle-Side-Angle postulate $\text{Δ}AED\cong \text{Δ}AFD$

Line $\overrightarrow{AM}$ is perpendicular bisector of $\overline{BC},$ implies point $A$ lies on the perpendicular bisector of $\overline{BC}$

Therefore, point $A$ is equidistant from endpoints of $\overline{BC},$ meaning $AB=AC$

  $\ln \text{Δ}AEB$ and $\text{Δ}AFC$

  $\angle AEB\cong \angle AFC\overrightarrow{AE}\perp \overrightarrow{BD};\overrightarrow{AF}\perp \overrightarrow{DF}$

  $\overrightarrow{AE}\cong \overrightarrow{AF}(\text{Δ}AED\cong \text{Δ}AFD,$ Corresponding parts of $\cong \text{Δs}$ $\cong )$

  $\overrightarrow{AB}\cong \overrightarrow{AC}AB=AC$

Therefore, by Angle-Side-Side postulate $\text{Δ}AEB\cong \text{Δ}AFB$

This gives $\overline{BE}\cong \overline{CF}$ since $\text{Δ}AEB\cong \text{Δ}AFB$ Corresponding parts of congruent triangles.

Hence, $\overline{BE}\cong \overline{CF}$ by Angle-Side-Side postulate.