Chapter 4.5, Problem 17PPS

To determine

To prove:

  $\overline{CD}\cong \overline{CB}$ for the following figure in the form of the paragraph:

  

Explanation of Solution

Given information:

  $C$ is the midpoint of $\overline{AE}$

  $\begin{array}{l}\overline{AE}\perp \overline{DE}\\ \overline{EA}\perp \overline{AB}\end{array}$

Proof:

From the given figure:

  

Know that $\overline{AE}$ and $\overline{EA}$ are perpendicular to $\overline{DE}$ and $\overline{AB}$ respectively and midpoint of two triangles is $C$ . The measure of angle $CED$ is $90°$ i.e. since $\overline{EA}$ perpendicular to $\overline{AB}$ . Since all right angles are congruent $\angle CED=\angle BAC$ . By using the mid-point theorem, get that $\overline{AC}\cong \overline{CE}$ . Since $\angle ECD$ and $\angle ACB$ are vertical angles $\angle ECD\cong \angle ACB$ . Use the angle side angle (ASA) congruence postulate gets that $\Delta CED\cong \Delta CAB$ .Know that corresponding parts of congruent triangles are congruent so, $\overline{CD}\cong \overline{CB}$

Hence proved.