Diagonals in a rectangle

Meet the diagonals of the rectangle:

Diagonal $AC$ and diagonal $DB$ are the diagonals of the rectangle shown.
$E$ is the intersection point of the diagonals.
Properties of rectangle diagonals:

1. The diagonals of a rectangle are equal to each other.
   That is - $AC=DB$
   You can use this property without proving it. If you are asked to prove this statement - you can easily do so by proving the congruence of triangles $ADC$ and $ABC$ using the Side-Angle-Side (SAS) postulate. After proving that the triangles are congruent, you can conclude that $AC = BD$, and therefore these are the diagonals of the rectangle and they are equal.
2. The diagonals of a rectangle bisect each other.
   That is - each diagonal divides the other diagonal into two equal parts so that all halves are equal.
   In fact: $AE=DE=CE=DE$
   Note - this property can help you in exercises where you need to prove that opposite triangles in a rectangle are congruent to each other. Additionally, this property can help you in exercises where you need to prove that the areas of all 4 triangles are identical.

Important to note -
The diagonals of a rectangle do intersect each other but:
- They are not perpendicular to each other - they don't form a right angle at their intersection.
- And they do not bisect the angles of the rectangle!

Now that you have learned everything you need to know about the diagonals of a rectangle, it's time to practice!

Practice:

1. Here is a rectangle.
a. Prove that triangle $AEB$ is equal in area to triangle $DEC$.
b. Is angle $DEC$ equal to $90$? Explain
c. Is angle $ABE$ equal to angle $CBE$? Explain

Solution:
a. Given a rectangle.
In a rectangle, the diagonals are equal and bisect each other, and from this we can conclude that:
$AE=BE=CE=DE$
We can prove congruence between triangle $AEB$and $DEC$
Side $AE=DE$ – explained above
Angle $AEB$= angle $DEC$ – alternate angles are equal
Side – BE=CE – explained above
Therefore, by SAS, triangle $AEB$ is congruent to triangle $DEC$
Therefore, we can conclude that the areas of the triangles are identical. Congruent triangles have identical areas.

b. Angle $DEC$ is not equal to $90$ degrees since the diagonals in the parallelogram are not perpendicular to each other.
c. Angle $ABE$ is not equal to angle $CBE$ since the diagonals in the triangle do not serve as angle bisectors.

Additional exercise:

Here is a rectangle.
a. Prove that triangle $ADE$ is congruent to triangle $BEC$
b. Prove that all four triangles formed in the rectangle have equal areas.

Solution:
a. Given a rectangle. In a rectangle, the diagonals are equal to each other and bisect each other, therefore:
$AE=BE=DE=CE$
We can prove congruence using side-angle-side:
Side – $AE=BE$ explained above
Angle $BEC$ equals angle $AED$ vertical angles are equal.
Side $DE=CE$ explained above
Therefore triangle $ADE$ is congruent to triangle $BEC$

b. To calculate area, let's recall the triangle area formula:
Multiply height to side by side and divide the result by $2$.

To calculate the area of triangle $ADE$ we need to draw a height to side $DE$
Let's call it $AG$ and mark it with a dashed line.

Note that the height $AG$ also serves as an external height for triangle $AEB$ - creating a right angle with the extension of side $BE$.
To calculate the area of triangle $AEG$, we will use the height $AG$ multiplied by $EB$ and divide the result by $2$.

We can deduce that the only different variables in calculating the two areas are $DE$ and $EB$ which are equal according to the property that diagonals in a rectangle bisect each other, therefore the areas of these two triangles are equal.
From here we need to prove congruence between triangle $AEG$ and triangle $BEC$ as we proved in part a. - two congruent triangles have identical areas.
And we need to prove congruence of triangle $AEB$ and triangle $DEC$ using side-angle-side.
Therefore, all four triangles in the rectangle have identical areas.