Question

If two sides of a cyclic quadrilateral are parallel, prove that the remaining two sides equal and the diagonals are also equal.

Answer 1

Let $ABCD$ be the cyclic quadrilateral with $AD$ parallel to $BC.$ The center of the circle is $O.$

Join $AC$

Now, since $AD$ is parallel to $BC$ and $AC$ is the transverse

$\angle DAC=\angle ACB$ (interior opposite angles)

$\angle DAC$ is subtended by chord $DC$ on the circumference and $\angle ACB$ is subtended by chord AB.

If the angles subtended by $2$ chords on the circumference of the circle are equal, then the lengths of the chords are also equal.

Hence, $DC=AB$ i.e. the $2$ remaining sides of the cyclic quadrilateral are equal.

Now, $BD$ and $AC$ are the two diagonals.

In $△ABD$ and $△DCA,$

$AB=CD$ $($proved above$)$

$\angle DCA=\angle DAB$ $($angles subtended by the same chord $AD)$

$\angle ACB=\angle DAC$ $($proved above$)$

Hence, $△ABD$ is congruent to $△DCA($by $ASA)$

Hence, $BD=AC$ (by CPCT)