Chapter 5, Problem 29RP

To determine

Prove that $ABCD$ is a rectangle.

Answer to Problem 29RP

Proved, $ABCD$ is a rectangle.

Explanation of Solution

Given information:

  

  $\Delta AEDand\Delta BEC$ are isosceles with congruent bases $\overline{AD}and\overline{BC}$

Calculation:

Consider the given figure,

  

It is given $\Delta AEDand\Delta BEC$ are isosceles with congruent bases $\overline{AD}and\overline{BC}$ ,

  $\begin{array}{l}\angle AED\cong \angle BEC\left[verticalanglearecongruent\right]\\ \angle EAD\cong \angle EDA,\angle EBC\cong \angle ECB\left[baseanglesarecongruent\right]\end{array}$

As per transitivity

  $\begin{array}{l}m\angle EAD=m\angle EDA=\frac{180-m\angle AED}{2}\\ m\angle EBC=m\angle ECB=\frac{180-m\angle BEC}{2}\\ \Rightarrow \angle EAD\cong \angle EDA\cong \angle EBC\cong \angle ECB\\ \overline{AD}\left|\right|\overline{BC}\left[alternate\mathrm{int}eriorangles\right]\end{array}$

  $ABCD$ is parallelogram,

Sides corresponding to congruent angles in an triangle are congruent,

  $\overline{AE}\cong \overline{DE},\overline{BE}\cong \overline{CE}$

  $\begin{array}{l}AC=AE+CE\\ BD=DE+BE\\ \overline{AC}\cong \overline{BD}\end{array}$

  $ABCD$ is rectangle.