Chapter 3.8, Problem 12PSB

To determine

The congruency of $\overrightarrow{JD}\cong \overrightarrow{HE}$ is to be proved

Explanation of Solution

Given information : The following information has been given

  $\overrightarrow{CD}\cong \overrightarrow{EF}$

  $\overrightarrow{JD}\perp \overrightarrow{JF}$

  $\overrightarrow{CH}\perp \overrightarrow{HE}$

  $\overrightarrow{CH}\cong \overrightarrow{JF}$

Formula used : Pythagoras theorem is used, with states that in right angled triangle ABC, if AB is the hypotenuse, then

  $a^2+b^2=c^2$

Proof : In the triangles FDJ and CEH, we can say that

  $\angle CHE\cong \angle FJD={90}^0$

  $\overrightarrow{CH}\cong \overrightarrow{JF}$

  $\overrightarrow{CE}\cong \overrightarrow{FD}$ (adding DE on both sides of $\overrightarrow{CD}\cong \overrightarrow{EF}$ )

Hence, now we use Pythagoras theorem on triangles FDJ and CEH. Thus, we get

  $\overrightarrow{JD}\cong \overrightarrow{HE}$

Hence, the congruency of $\overrightarrow{JD}\cong \overrightarrow{HE}$ is proved