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Big B's theorem (Posted on 2012-10-14) Difficulty: 3 of 5
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

No Solution Yet Submitted by Ady TZIDON    
Rating: 2.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 2 of 2 |

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 2012-10-14 19:04:43
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