Brahmagupta's Theorem: What is it?
A Mathematical Droodle

Explanation

Brahmagupta's Theorem

In a cyclic quadrilateral having perpendicular diagonals, the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

There are several right angles: DET in EDT, AET in AET, DTA in ADT, BTC in BCT.

From the first three triangles, we have

(1)	DTE = EAT and ETA = EDT.

We also have two pairs of vertically opposite angles:

(2)	DTE = BTQ and ETA = CTQ.

Chords AB and DC subtend pairs of angles:

(3)	ADB = ACB and DAC = DBC.

By comparing (1)-(3) we conclude that

(4)	TCQ = CTQ and BTQ = TBQ.

Both triangles CQT and BQT are isosceles and BQ = QT = CQ.

The theorem of course admits the following variation:

In a cyclic quadrilateral having perpendicular diagonals, the perpendicular from the midpoint of a side to the opposite side passes through the point of intersection of the diagonals.

There are four such perpendiculars and all four pass through the point of intersection of the diagonals. In other words, the four perpendiculars from the midpoints of the sides to the opposite side are concurrent, and the point of concurrency coincides with the intersection of the diagonals.

Now, all this is true under the condition of orthogonality of the diagonals. Orthogonality plays an important role in both the formulation and the proof of the theorem. It's therefore a curiosity that the theorem admits a generalization that does not require the diagonals to be orthogonal. In the more general case the four lines are still concurrent, but they no longer meet at the intersection of the diagonals.

References

1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.