Question

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer 1

Given: $\square ABCD$ is a quadrilateral.
diag $AC=$ diag $BD$, intersecting at $E$.
$AC$ and $BD$ are perpendicular bisectors of each other.
$\therefore \angle E={90}^o$
To prove: $\square ABCD$ is a square.
Solution:
A square is a parallelogram with all sides equal and one angle is ${90}^{\circ }$.
First let us prove all sides are equal.
In $△ABE$ and $△ADE$.
$BE=DE$ ....given
$AE=AE$ ...common side
$\angle AEB\cong \angle AED$ ....each ${90}^o$

$\therefore △ABE\cong △ADE$ ...SAS test of congruence

$\therefore AB=AD$ ...corrseponding sides of congruent triangles(c.s.c.t).......(1)
Similarly, we can prove $△ABE\cong △CBE$

$\therefore AB=CB$ ....c.s.c.t. ....(2)

And, from $△ADE\cong △CDE$

$\therefore AD=CD$ ....c.s.c.t. ....(3)

$\therefore$ From (1), (3) and (4),

$AB=CB=CD=AD$ ............(4)
Therefore, all sides are equal.

Now, we prove one angle is ${90}^{\circ }$
In $\Delta ABC$ and $\Delta ABD$
$AC=BD$ ........given
$BC=AD$ ........ Opposite sides of parallelogram are equal.
$AB=AB$ .......common side
Therefore $\Delta ACB\cong \Delta BDA$ ...... By SSS of congruence.
So, $\angle ABC=\angle BAD$ .......Corresponding angles of congruent triangles (c.a.c.t)
Since, $AD||BC$ and $AB$ is transversal
So, $\angle A+\angle B={180}^{\circ }$ .....interion angles on same side of a transversal is supplementary .
Since $\angle A=\angle B$
So, $\angle A+\angle A={180}^{\circ }$
$2A=180$
$A=\frac{180}{2}$
$\therefore A={90}^{\circ }$
Thus, in quad.$ABCD$
$AB=CB=CD=AD$ and $\angle A={90}^{\circ }$

$\therefore \square ABCD$ is a square ....By definition