Cyclic Quadrilaterals from Isosceles Triangles

As an extension to an earlier post, let's add a condition that the quadrilateral must also be cyclic.

 

Solution 

If the triangles are connected base-to-base, like in Setup 1, the symmetry forces it to be a kite, with \(x+y=90\). A special case is if the other angles are also \(90^{\circ}\) in which case it becomes a square.

If the triangles are set up base-to-leg, like in Setup 2, the apex angle in the left triangle must be \(180-2y\) and because opposite angles in a cyclic quadrilateral sum to \(180\), this means that \(x=2y\). The apex angles in the right triangle is now \(180-4y\).

What quadrilateral could this be?

First off, the apex angle of the left triangle is \(180-2y\) and this is opposite the angle of size \(x\) in the cyclic quadrilateral. They sum to \(180\) so \(180-2y+x=180 \implies x=2y\).

Now, let's use this fact to test types of quadrilateral. It cannot be a square or rectangle as \(x\) cannot be \(90\).

Can it be a kite? If it were, either \(x=180-2y\) and recalling the fact that \(x=2y\), we get \(x=90\) which is not possible. Or, the other two angles are equal giving \(y+(180-2x)=y+x\implies 180=3x \implies x=60\). This makes the triangle on the right an equilateral triangle and \(y=30\). The kite then has angles 120, 90, 60 and 90. This seems feasible.

Can it be a trapezium? Assume it is, then we can use facts about angles in parallel lines. Either \(y=x\), because of alternate angles, but then \(y=x=2y \implies y=0\) which is not possible. Considering a different pair of lines to be parallel, we get a different pair of alternate angles producing the equation \(y=180-2x\) and recalling the earlier finding that \(x=2y\) we get \(y=36\). The trapezium would have angles 108, 108, 72 and 72.

Overall, the quadrilateral can be a square, kite or an isosceles trapezium.