Question

Prove that if the diagonals of a parallelogram are congruent, then it must be a rectangle. (Hint: you might need to use the fact that equilateral angles add up to 360°, or it’s also possible to do it with triangle sum or EAT. Both of these methods are recommended.)

Answer 1

We will have the following:

We will have the paralelogram ABCD and it has the diagonals AC congruent with BD.

Now, we will have that:

*Segment AB is parallel with segment CD [Given by definition]

*Segment AC is parallel with segment BD [Given by definition]

*We will have that the segment AD will be congruent with segment BC. [Opposite sides of a parallelogram are congruent]

*We then will have that segments AB will be congruent with AB, and segment DC will be congruent with DC. [Reflexive property.

*Then we will have:

$\begin{cases}\Delta DAB\cong\Delta CBA \\ \\ \Delta ADC\cong BCD\end{cases}$

[We will have this by SSS congruence]

*Then:

$\begin{cases}<\text{DAB}\cong[This is directly derived from the previous point]

*Then we will have:

[tex]\begin{cases}<\text{DAB}+[This is that the angles DAB & CBA are supplementary, likewise angles ADC and BCD]

*And since congruent angles are equal we will have that:

[tex]\begin{cases}[Substitution]

*Then:

[tex]\begin{cases}2<\text{DAB}=180 \\ \\ 2<\text{ADC}=180\end{cases}$

[Addition]

*Then we determine the value of each angle:

$\begin{cases}<\text{DAB}=90 \\ \\ <\text{ADC}=90\end{cases}$

[Division]

*Then we will have that:

*Then we have proved that a parallelogram with congruent diagonals is a rectangle.