If two non parallel sides of trapezium are equal. Prove that it is cyclic If the two non parallel sides of a trapezium are equal then prove that it is cyclic. I know that if the two sides are parallel the trapezium will become a isoceles trapezium. _But how to prove that it always will?_ The quadrilateral should not be a parallelogram i.e. the sides should remain non parallel And is there a visual proof without words

Let the trapezium $ABCD$ be isosceles and the parallel sides be $AB$ and $CD$. The equal length non-parallel sides are $BC$, $DA$. The angles $\angle DAB$ and $\angle CDA$ are supplementary in the parallel lines so $\angle DAB + \angle CDA = \pi$.

The trapezium is symmetric so the angles $\angle DAB$ and $\angle ABC$ are equal.

So we have $\angle ABC + \angle CDA = \pi$, i.e. we have opposite angles in the trapezium add up to $\pi$. This is necessary and sufficient for a quadrilateral to be cyclic.

For a visual illustration, draw a isosceles trapezium, then draw the perpendicular bisectors of the two non-parallel sides. Think about the distance from the point where the bisectors intersect and each of the vertices of the trapezium.