# Brahmagupta Theorem

## Theorem

If a cyclic quadrilateral has diagonals which are perpendicular, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

Specifically:

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are perpendicular, crossing at $M$.

Let $EF$ be a line passing through $M$ and crossing opposite sides $BC$ and $AD$ of $ABCD$.

Then $EF$ is perpendicular to $BC$ if and only if $F$ is the midpoint of $AD$.

## Proof

### Sufficient Condition

Suppose that $EF$ is perpendicular to $BC$.

We need to prove that $AF = FD$.

Thus:

\(\ds \angle FAM\) | \(=\) | \(\ds \angle CAD\) | producing $AF$ to $D$ and $AM$ to $C$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle CBD\) | Angles in Same Segment of Circle are Equal: both subtend $CD$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle CBM\) | producing $BM$ to $D$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle CME\) | both are complements of $\angle BCM$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle FMA\) | Vertical Angle Theorem |

Then by Triangle with Two Equal Angles is Isosceles, it follows that $AF = FM$.

Similarly:

\(\ds \angle FDM\) | \(=\) | \(\ds \angle ADB\) | producing $DF$ to $A$ and $DM$ to $B$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle ACB\) | Angles in Same Segment of Circle are Equal: both subtend $AB$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle BCM\) | producing $BM$ to $A$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle BME\) | both are complements to $\angle CBM$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle DMF\) | Vertical Angle Theorem |

Then by Triangle with Two Equal Angles is Isosceles, it follows that $FD = FM$.

So $AF = FD$, as we needed to show.

$\Box$

### Necessary Condition

Now suppose that $AF = FD$.

We now need to show that $EF$ is perpendicular to $BC$.

From Thales' Theorem (indirectly) we have that $AF = FM = FD$.

So:

\(\ds \angle EBM\) | \(=\) | \(\ds \angle CBD\) | producing $EB$ to $C$ and $BM$ to $D$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle CAD\) | Angles in Same Segment of Circle are Equal: both subtend $CD$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle FAM\) | producing $AM$ to $C$ and $FA$ to $D$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle AMF\) | Isosceles Triangle has Two Equal Angles, and $AF = FM$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \angle EMC\) | Vertical Angle Theorem |

We note the result Sum of Angles of Triangle equals Two Right Angles.

We have that $\angle EBM$ and $\angle ECM$ are complementary, as both are angles in $\triangle CBM$, which is a right triangle.

So $\angle EMC$ and $\angle ECM$ are complementary, which means that $\angle CEM$ must be a right angle.

Hence by definition $EF$ is perpendicular to $BC$, as we were to show.

$\blacksquare$

## Source of Name

This entry was named for Brahmagupta.