euclidean-geometrygeometry
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
You can parametrize the trapezium via its height $h$. Then the longer side is $$ 2 \sqrt{k^2-h^2} + k $$ and its area is $$ h (\sqrt{k^2-h^2} + k)\ .$$ The derivative for $h$ is $$k - \frac{h^2}{\sqrt{-h^2 + k^2}} + \sqrt{-h^2 + k^2}\ ,$$ which is zero for $h=0$ or $h=\pm \frac{\sqrt{3}}{2}k$. $h=\frac{\sqrt{3}}{2}k$ gives the proposed optimum, which you can be sure about after also checking the border case $h=k$ (which gives an area of $k^2$).
Perhaps it is not surprising that all the segments (not only those that are actually depicted) going through the midpoint of the red line are halved by the same red point:
Taking any trapezium the two diagonals will be parallel to two of the colored lines above. That is, their midpoints will be equidistant to both of the black parallel lines...
Best Answer
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.