Question

Show that the diagonals of the square are equal and bisect each other at right angles.

Answer 1

Let a square $ABCD$,

$AC$ and $BD$ are the diagonals intersects at $E$.

To prove:

(i) $AC=BD$

(II) $AC$ and $BD$ bisects each other at right angles.

Proof:

Consider triangles $ADC$ and $BDC$

$AD=BC$

$DC=DC$

$\angle ADC=\angle BCD$

with SAS property triangles $\Delta ADC$ and $\Delta BCD$ are congruent

so $AC=BD$ ---(i)

Now consider triangle $ADC$

$AD=DC$

So, $\angle DAC$ and $\angle DCA$ are equal

but angle $\angle ADC={90}^{\circ }$

So, angle $\angle DAC=\frac{{90}^{\circ }}{2}={45}^{\circ }$

in the similar way $\angle ADB={45}^{\circ }$

now consider triangle $ADE$

$\angle DAE+\angle ADE+\angle AED={180}^{\circ }$

$\Rightarrow {45}^{\circ }+{45}^{\circ }+\angle AED={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle AED={180}^{\circ }$

$\Rightarrow \angle AED={180}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle AED={90}^{\circ }$

$\Rightarrow \angle AED=\angle AEB=\angle CEB=\angle CED{90}^{\circ }$

Hence, $AC\perp BD$.---(ii)

Now, lets take $\Delta AEB$ and $\Delta AED$

$\angle AED=\angle AEB={90}^{\circ }$ [Since, proved]

Hypotenuse $AD=$ Hypotenuse $AB$ [Since, $ABCD$ is a square]

$AE=AE$ [Since, common side]

By RHS congruence rule,

$\Delta AED\cong \Delta AEB$

By C.P.C.T,

$DE=BE$---(a)

Similarly, $\Delta AED$ and $\Delta CED$

$\angle AED=\angle CED={90}^{\circ }$ [Since, proved]

Hypotenuse $AD=$ Hypotenuse $CD$ [Since, $ABCD$ is a square]

$DE=DE$ [Since, common side]

By RHS congruence rule, $\Delta AED\cong \Delta CED$

By C.P.C.T,

$AE=CE$---(b)

From (a) and (b) diagonals of the square $ABCD$ bisect each other at the point $E$.---(iii)

From (i), (ii) and (iii), the diagonals of the square are equal and bisect each other at right angles