Chapter 5.3, Problem 29PSD

To determine

To Prove: $\overline{BE}=\frac{1}{2}\left(\overline{CD}\right)$

Explanation of Solution

Given:

  

  $\begin{array}{l}\overleftrightarrow{BE}\parallel \overleftrightarrow{CD}\text{and }\overline{BE}\cong \overline{EF}\\ \text{B is the midpoint of }\overline{AC}\\ \text{E is the midpoint of }\overline{AD}\end{array}$

Concept Used:

When two parallel lines are intersected by a transversal corresponding angles are equal.

Proof:

Join F and D, we get

  

  $\begin{array}{l}\overleftrightarrow{BE}\parallel \overleftrightarrow{CD}\text{and }\overline{BE}\cong \overline{EF}\\ \overline{BF}\parallel \overline{CD}\\ \text{B is the midpoint of }\overline{AC}\\ \text{E is the midpoint of }\overline{AD}\\ \text{In }△\text{AEB and }△\text{DEF}\\ \overline{AE}\cong \overline{DE}\\ \angle AEB\cong \angle DEF\\ \overline{EB}\cong \overline{EF}\\ \therefore △\text{AEB }\cong △\text{DEF}\\ \Rightarrow \angle ABE\cong \angle DFE\\ \Rightarrow \overline{AB}\parallel \overline{DF}\\ \Rightarrow \overline{BC}\parallel \overline{DF}\\ \Rightarrow BCDF\text{is a parallelogram}\text{.}\\ \Rightarrow \overline{BF}\cong \overline{CD}\\ \because \overline{BE}\cong \overline{EF}\\ \Rightarrow \overline{BE}=\frac{1}{2}\left(\overline{CD}\right)\end{array}$