Chapter 6.5, Problem 14WE

To determine

To prove: $DK>DJ$ in two column form.

Explanation of Solution

Given information:

Given figure,

Given:

$\overline{DE}\perp$ Plane $M$ ; $EK>EJ$

Calculation:

Consider the following diagram:

Consider the figure provided in the textbook.

Also Consider that $\overline{DE}\perp$ Plane $M$ ; $EK>EJ$

Prove that $DK>DJ$ as follows:

Use Pythagoras theorem,

  $\begin{array}{l}D{K}^2=D{E}^2+E{K}^2\\ D{J}^2=D{E}^2+E{J}^2\end{array}$

Since,

  $\begin{array}{c}EK>EJ\\ E{K}^2>E{J}^2\\ D{E}^2+E{K}^2>D{E}^2+E{J}^2\\ D{K}^2>D{J}^2\end{array}$

Implies $DK>DJ$

Thus, the inequality $DK>DJ$ has been proved.