Brahmagupta Theorem

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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

BrahmaguptaTheorem.png

Sufficient Condition

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

We need to prove that $AF = FD$.

Thus:

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

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


Similarly:

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

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