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
Let E be the intersection of the diagonals
AC and BD. Let F be the point on side CD
such that line EF is perpendicular to side
CD. Let G be the intersection of line EF
and side AB. We need to prove that
GA = GB.
<GAE = <BAC  Same angle
= <BDC  ABCD is concylic
= <EDF  Same angle
= <CEF  Similar rt. triangles
= <GEA  Vertical angles
Therefore, GA = GE.
<GBE = <ABD  Same angle
= <ACD  ABCD is concylic
= <ECF  Same angle
= <DEF  Similar rt. triangles
= <GEB  Vertical angles
Therefore, GB = GE.
Hence, GA = GB.
QED

Posted by Bractals
on 20121014 19:04:43 