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
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Posted by Bractals
on 2012-10-14 19:04:43 |
Solution
