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 $BC$ $F$ is the midpoint of $AD$.

Proof

 * BrahmaguptaTheorem.png

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

We need to prove that $AF = FD$.

Thus:

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

Similarly:

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

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

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:

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.