Quadrilateral is Cyclic iff Opposite Angles sum to Two Right Angles

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Theorem

Given $\Box ABCD$ with $A, B$ and $D$ on a circle.

Let $\angle ABC$ and $\angle ADC$ add to two right angles.

Then $C$ lies on the circle, and $\Box ABCD$ is a cyclic quadrilateral.

Proof

Sufficient Condition

By Sum of Internal Angles of Polygon, since $\Box ABCD$ is a quadrilateral:

$\angle ABC + \angle BCD + \angle BAD + \angle ADC = 360$

Therefore $\angle BAD$ and $\angle BCD$ are also supplementary angles.

Suppose $C$ does not lie on the circle, but lies internal.

Extend $DC$ to meet the circle at $C'$.

By Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:

$\angle BC'D$ is supplementary to $\angle BAD$.

It follows that:

$\angle BC'D = \angle BCD$

By External Angle of Triangle equals Sum of other Internal Angles:

$\angle BCD = \angle BC'D + \angle C'BC$

Hence $BCD > BC'D$.

But this is a contradiction.

Suppose $C$ lies outside the circle.

Draw $DC$ and find the point $C'$ where $DC$ meets the circle.

By External Angle of Triangle equals Sum of other Internal Angles:

$\angle BC'D = \angle BCD + \angle C'BC$

Hence $BC'D > BCD$.

But this is a contradiction.

Since $C$ is neither internal to the circle nor external to it, it must lie on the circle.

$\Box$

Necessary Condition

This follows by Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles.

$\blacksquare$