**Brahmagupta's theorem** is a result in

geometry. It states that if a

cyclic quadrilateral is

orthodiagonal (that is, has

perpendicular diagonals), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. It is named after the

Indian mathematician Brahmagupta.

More specifically, let

*A*,

*B*,

*C* and

*D* be four points on a circle such that the lines

*AC* and

*BD* are perpendicular. Denote the intersection of

*AC* and

*BD* by

*M*. Drop the perpendicular from

*M* to the line

*BC*, calling the intersection

*E*. Let

*F* be the intersection of the line

*EM* and the edge

*AD*. Then, the theorem states that

*F* is the midpoint

*AD*.

## Proof

We need to prove that

*AF* =

*FD*. We will prove that both

*AF* and

*FD* are in fact equal to

*FM*.

To prove that

*AF* =

*FM*, first note that the angles

*FAM* and

*CBM* are equal, because they are

inscribed angles that intercept the same arc of the circle. Furthermore, the angles

*CBM* and

*CME* are both

complementary to angle

*BCM* (i.e., they add up to 90°), and are therefore equal. Finally, the angles

*CME* and

*FMA* are the same. Hence,

*AFM* is an

isosceles triangle, and thus the sides

*AF* and

*FM* are equal.

The proof that

*FD* =

*FM* goes similarly: the angles

*FDM*,

*BCM*,

*BME* and

*DMF* are all equal, so

*DFM* is an isosceles triangle, so

*FD* =

*FM*. It follows that

*AF* =

*FD*, as the theorem claims.

## See also

## External links

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