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Show that the diagonals of a square are equal and bisect each other at right angles.
To do :
We have to show that the diagonals of a square are equal and bisect each other at right angles.
Solution :
Let $ABCD$ be a square in which diagonals $AC$ and $BD$ intersect each other at $O$.
In $\triangle ABC$ and $\triangle BAD$,
$AB=AB$ (Common side)
$BC=AD$ (All the sides of a square are equal)
$\angle ABC= \angle BAD=90^o$
Therefore, by SAS congruency, we get,
$\triangle ABC \cong \triangle BAD$
So, $AC=BD$ (CPCT)
The diagonals are equal.
In $\triangle AOB$ and $\triangle COD$,
$\angle BAO = \angle DCO$ (Alternate interior angles are equal)
$\angle AOB = \angle COD$ (Vertically opposite angles are equal)
$AB = CD$
Therefore, by AAS congruency, we get,
$\triangle AOB \cong \triangle COD$
So,
$AO = CO$ (CPCT)
This implies,
Diagonal bisect each other.
In $\triangle AOB$ and $\triangle COB$,
$OB = OB$ (Common side)
$AO = CO$ (Diagonals bisect each other)
$AB = BC$ (Sides of the square are equal)
Therefore, by SSS congruency, we get,
$\triangle AOB \cong \triangle COB$
This implies,
$\angle AOB = \angle COB$
$\angle AOB+\angle COB = 180^o$ (Linear pair)
This implies,
$\angle AOB+\angle AOB =180^o$
$\angle AOB=\frac{180^o}{2}$
$\angle AOB=90^o$
$\angle COB =\angle AOB= 90^o$
Diagonals are equal and bisect each other at right angles.
Therefore,
The diagonals of a square are equal and bisect each other at right angles.
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