Question 934674
If {{{AD = CB}}} then the trapezoid {{{ABCD}}} is a {{{cyclic-quadrilateral}}}.


Consider a trapezium {{{ABCD}}} with {{{AB}}} ||{{{CD}}} and {{{AD=BC}}}
Draw AM perpendicular to CD and BN  perpendicular to CD.

In triangle {{{AMD}}} and triangle {{{BNC}}},
{{{AD = BC}}}..................................................... (Given)
< {{{AMD}}} = < {{{BNC}}}........................................ (By construction, each is {{{90}}}°)
{{{AM = BM}}}.................................................... (Perpendicular distance between two parallel lines is same)
 &#916; {{{AMD}}} congruent &#916;{{{BNC}}}............................................... (RHS congruence rule)
< {{{ADC}}} = < {{{BCD}}}................................................. (CPCT) ... (1)
< {{{BAD}}} and < {{{ADC}}}.............................................. are on the same side of transversal AD.
< {{{BAD}}} + <{{{ ADC}}} = {{{180}}}° ...................................... (2)
< {{{BAD}}} + < {{{BCD}}} = {{{180}}}°.................................... [Using equation (1)]

This equation shows that the opposite angles are supplementary.
Therefore, {{{ABCD}}} is a {{{cyclic}}}{{{ quadrilateral}}}.