Question

In the given figure, $ABCD$ is a quadrilateral in which $AD=BC$ and $\angle ADC=\angle BCD$. Show that the points $A,B,C,D$ lie on a circle.

Answer 1

Given:

$ABCD$ is a quadrilateral in which $AD=BC$.

$\angle ADC=\angle BCD$ .

To prove:

The points $A,B,C,D$ lie on a circle.

Construction:

Draw $DE\perp AB$ and $CF\perp AB$.

Proof:

From the construction, $DE\perp AB$ and $CF\perp AB$

$\Rightarrow \angle DEA={90}^{\circ }$ and $\angle CFE={90}^{\circ }$

i.e., $DE\parallel CF$ [Since, corresponding angles are equal]

Here, $EF$ and $DC$ is the distance between the parallel sides.

So, $EF=DC$

$\Rightarrow \angle EDC={90}^{\circ }$ and $\angle FCD={90}^{\circ }$

Hence, $EFCD$ is a rectangle.

Now,

$\angle ADC=\angle ADE+{90}^{\circ }$

$\angle BCD=\angle BCF+{90}^{\circ }$

But, $\angle ADC=\angle BCD$

$\Rightarrow \angle ADE+{90}^{\circ }=\angle BCF+{90}^{\circ }$

$\Rightarrow \angle ADE=\angle BCF$ ---(1)

Now, in $\Delta ADE$ and $\Delta BCF$

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

$\angle ADE=\angle BCF$ [Proved]

$AD=BC$ [Given}

By AAS congruency,

$\Delta ADE\cong \Delta BCF$
Thus, the corresponding parts of congruent triangles are equal.
$\Rightarrow \angle A=\angle B$

and $\angle C=\angle D$ [Since, $\angle ADC=\angle BCD$]

We know that, In a quadrilateral, the sum of four angles would be ${360}^{\circ }$.
$\Rightarrow \angle A+\angle B+\angle C+\angle D={360}^{\circ }$
$\Rightarrow \angle B+\angle B+\angle D+\angle D=360$ (Since, $\angle A=\angle B$ and $\angle C=\angle D$)
$\Rightarrow 2\angle B+2\angle D=360$
$\Rightarrow \angle B+\angle D={180}^{\circ }$
Since, in cyclic quadrilateral sum of the opposite angles are supplementary.
So, $ABCD$ is a cyclic quadrilateral.

Hence, The points $A,B,C,D$ lie on a circle.