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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle FAM\)	\(=\)	\(\displaystyle \angle CAD\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by producing $AF$ to $D$ and $AM$ to $C$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle CBD\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Angles in Same Segment of Circle are Equal: both subtend $CD$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle CBM\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by producing $BM$ to $D$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle CME\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	both are complementary to $\angle BCM$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle FMA\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Vertical Angle Theorem

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

Similarly:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle FDM\)	\(=\)	\(\displaystyle \angle ADB\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by producing $DF$ to $A$ and $DM$ to $B$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle ACB\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Angles in Same Segment of Circle are Equal: both subtend $AB$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle BCM\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by producing $BM$ to $A$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle BME\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	both are complementary to $\angle CBM$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle DMF\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	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.

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle EBM\)	\(=\)	\(\displaystyle \angle CBD\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by producing $EB$ to $C$ and $BM$ to $D$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle CAD\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Angles in Same Segment of Circle are Equal: both subtend $CD$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle FAM\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by producing $AM$ to $C$ and $FA$ to $D$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle AMF\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Isosceles Triangles have Two Equal Angles, and $AF = FM$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \angle EMC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	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.