Brahmagupta*'s theorem states that if a cyclic quadrilateral ABCD has perpendicular diagonals, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.
Today a high school student can prove it.
How about you?
*Indian mathematician of 7th century
Start from the other end.
Construct perpendicular lines PQ, RS. Mark a point T in the plane and reflect about PQ and RS to produce the rectangle TT'T''T'''. Mark a point O inside the rectangle. Reflect O about two adjacent sides of TT'T''T''' to points A and B. Note the two kites so formed ATOT''' and BTOT'. By construction these kites are at right angles to each other and ATB is a straight line of which T is the centre point. Extend BT' to twice its length at C, and do the same with AT''' at D. to complete the quadrilateral ABCD.
This construction is not necessarily cyclic but if and only if it is, then:
(1) every perpendicular dropped from A,B,C, or D will be parallel to one of the opposite sides of the quadrilateral, and
(2) each adjacent pair of kites will share one angle that is the same, next to the opposite corners of the rectangle.
The foregoing construction can easily be shown to incorporate the key points of the given problem.
All we now need to know is:
(3) that if a figure is a a kite, then the angle, say x, of a perpendicular dropped from a vertex, A, to the opposite side, BC, to the diagonal AC, will be of exactly half the size of angle ABC.
Now by parallel lines (the diagonals of the quadrilateral and the sides of the rectangle) we have x equal to half the angle of the diagonally opposite vertex of the adjacent kite, namely 2x, but by (2) we also have angle ABC= 2x. So, by (3) the line from the midpoint of the side, passing through the point of intersection of the diagonals, intersects BC at a right angle, as was to be proved.
Posted by broll
on 2012-10-14 11:05:07