Chapter 5.1, Problem 12PSB

To determine

To Prove: $\overline{DE}\ne \overline{EF}$ .

Explanation of Solution

Given:

  

  $\overline{HE}\text{is not the perpendicular bisector of }\overline{DF}$

Concept Used:

Two line segments are said to be congruent if they are of the same measure.

Proof:

Join points D and F with the centre O of the circle.

  

Here, we have

  $\begin{array}{l}\text{In }\Delta ODE\text{and }\Delta OFE\\ \overline{OE}\cong \overline{OE}\left(\because \text{Common side}\right)\\ \overline{OE}\cong \overline{OE}\left(\because \text{Radii of circle}\right)\\ \Rightarrow \angle ODE\cong \angle OFE\left(\because \text{Interior opposite angles made by sides of equal measure are also equal}\right)\\ \Rightarrow \Delta ODE\cong \Delta OFE\\ \text{But,}\overline{DE}\ne \overline{EF}\left(\because \overline{HE}\text{ is not the }\perp \text{bisector of side }\overline{HE}\right)\end{array}$