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

 * [[Image:BrahmaguptaTheorem.png|500px]]

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.