Chapter 14, Problem 4PE

To determine

To prove: the diagonals of a rectangle intersect to form four congruent segment.

Explanation of Solution

Given information:

  $▭ABCD$ is a rectangle with diagonal $AC$ , and $BD$ .

  

Calculation:

We will show that $\Delta ABC$ , and $\Delta ABD$ are congruent so that its common sides are the diagonals are also congruent.

So, from $\Delta ABC$ , and $\Delta ABD$ ,

  $AB=AB$ (Common side)

  $\angle DAB=\angle ABC$ (angles corresponding to side $AB$ )

  $DA=CB$ (Opposite side are equal)

That is by side-angle-side rule it follows that, $\Delta ABC$ and $\Delta ABD$ are congruent. Moreover,

  $\Delta ABC=\Delta ABD$

Hence as a common sides both the diagonals $AC$ and $BD$ are also congruent.

Since $E$ is the mid-point of $AC,BD$ , therefore $AE,EC,BE,ED$ are congruent to each other.

Hence proved.