Question

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer 1

Hint: In this question it is given that we have if the diagonals of a parallelogram are equal, i.e, PR=QS, then we have to show that it is a rectangle. So for this we need to draw the diagram first,

So for the solution first we need to show that the $$\mathrm{△}PQR$$ and $$\mathrm{△}SRQ$$ are congruent to each other and to show that PQRS is a rectangle, we have to prove that one of its interior angles is ${90}^{\circ }$.
Complete step-by-step solution:
In $$\mathrm{△}PQR$$ and $$\mathrm{△}SRQ$$,
PQ = SR (Opposite sides of a parallelogram are equal)
QR = QR (Common side)
PR = SQ (Since the diagonals are equal)
Therefore, by SSS(Side-Side-Side) property we can say that,
$$\mathrm{△}PQR\cong \mathrm{△}SRQ$$
Now by CPCT, i.e, “if two or more triangles which are congruent to each other then the corresponding angles and the sides of the triangles are equal to each other”. So we can write,
$$\mathrm{\angle }PQR=\mathrm{\angle }SRQ$$.......(1)

Since adjacent angles of a parallelogram are supplementary. (Consecutive interior angles)
$$\mathrm{\angle }PQR+\mathrm{\angle }SRQ={180}^{\circ }$$
$$\Rightarrow \mathrm{\angle }PQR+\mathrm{\angle }PQR={180}^{\circ }$$
$$\Rightarrow 2\mathrm{\angle }PQR={180}^{\circ }$$
$$\Rightarrow \mathrm{\angle }PQR={\displaystyle \frac{{180}^{\circ }}{2}}$$
$$\Rightarrow \mathrm{\angle }PQR={90}^{\circ }$$
Since PQRS is a parallelogram and one of its interior angles is 90º, PQRS is a rectangle.
Note: To show that a parallelogram is a rectangle, you no need to show every angle is ${90}^{\circ }$, because if one angle is ${90}^{\circ }$ then it indirectly implies that every other angles is ${90}^{\circ }$.
So as we know that in a parallelogram the opposite angles are always equal, so from here you will get, if one angle is ${90}^{\circ }$ then it’s opposite angle is also ${90}^{\circ }$. Again we also know that the summation of the adjacent angles is ${180}^{\circ }$, so from here also you are able to find that the remaining two angels are also ${90}^{\circ }$.