Quadrilateral is Cyclic iff Opposite Angles sum to Two Right Angles
It has been suggested that this page be renamed. In particular: This is not an iff proof, it is merely a converse To discuss this page in more detail, feel free to use the talk page. |
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.
$\blacksquare$