Triangles and quadrilaterals inscribed in a circle

Theorem: Through three points that are not on one line passes one circle, and only one.

• The centre of a circle through $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$ is at equal distances from them, so it is on the perpendicular bisector of $\displaystyle AB$ and on that of $\displaystyle BC$.
• Two perpendicular bisectors of a triangle’s sides meet at one point – the meeting point of the perpendicular bisectors: there is such a centre, and only one. The radius is its distance from $\displaystyle A$.
• Three points on one line have no such circle: the perpendicular bisectors of $\displaystyle AB$ and $\displaystyle BC$ are then parallel.

So every triangle is inscribed in a circle – its circumscribed circle – and a polygon is inscribed in a circle when all its vertices are on the circle.

Theorem: In a quadrilateral inscribed in a circle, every two opposite angles add up to $\displaystyle 180^{\circ}$.

Given: the quadrilateral $\displaystyle ABCD$ is inscribed in a circle.
To prove: $\displaystyle {\angle BAD+\angle BCD}={180^{\circ}}$.

The proof below draws the diagonals. The angle $\displaystyle A$ splits into two inscribed angles, each equal to an angle of the triangle $\displaystyle BCD$:

The angles $\displaystyle B$ and $\displaystyle D$ add up to $\displaystyle 180^{\circ}$ too: the four angles of a quadrilateral add up to $\displaystyle 360^{\circ}$.

Theorem: A quadrilateral in which two opposite angles add up to $\displaystyle 180^{\circ}$ can be inscribed in a circle.

• Its proof is by contradiction: draw the circle through $\displaystyle A$, $\displaystyle B$ and $\displaystyle D$, and assume $\displaystyle C$ is not on it. The line $\displaystyle DC$ meets the circle at another point $\displaystyle C'$; $\displaystyle ABC'D$ is inscribed, so $\displaystyle {\angle BC'D}={180^{\circ}-\angle A}={\angle BCD}$. Then in the triangle $\displaystyle BCC'$ an exterior angle equals an interior angle not next to it – which cannot be.
• So these quadrilaterals can be inscribed in a circle: every rectangle (each pair of opposite angles is $\displaystyle 90^{\circ}+90^{\circ}$) and every isosceles trapezoid (the angles at one base are equal, and the angles beside a leg add up to $\displaystyle 180^{\circ}$).
• A parallelogram that is not a rectangle cannot: its opposite angles are equal, and not $\displaystyle 90^{\circ}$ each.

The quadrilateral $\displaystyle ABCD$ is inscribed in a circle.

An exterior angle of a quadrilateral inscribed in a circle equals the interior angle opposite it.
$\displaystyle ABCD$ is inscribed in a circle, and $\displaystyle E$ is on the extension of $\displaystyle AB$ beyond $\displaystyle B$.
Prove: $\displaystyle {\angle CBE}={\angle ADC}$.

The trapezoid $\displaystyle ABCD$ ($\displaystyle AB\parallel DC$) is inscribed in a circle.
Prove: $\displaystyle {AD}={BC}$ – the trapezoid is isosceles.
(A hint: draw the diagonal $\displaystyle AC$, and look at the arcs $\displaystyle AD$ and $\displaystyle BC$.)

The quadrilateral $\displaystyle ABCD$ is inscribed in a circle, and its diagonal $\displaystyle AC$ is a diameter.
$\displaystyle {\angle BAC}={25^{\circ}}$ and $\displaystyle {\angle CAD}={40^{\circ}}$.

A kite has two equal opposite angles, between its unequal sides.
When can a kite be inscribed in a circle?
Explain.