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Equal Segments (Posted on 2013-03-20) Difficulty: 3 of 5

Let M be the circumcenter of ΔABC ( ∠B = 90° ),
    E the point on altitude BD of ΔABC such that 
      ∠AEC = 135°,
    F the point such that BDCF is a rectangle. and
    G the intersection of line segment MF and the
      circumcircle of ΔABC.

Prove that |DE| = |FG|.

See The Solution Submitted by Bractals    
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Solution Different approach Comment 2 of 2 |

Produce CE to meet the circle at P and BD to meet the circle at T.

Triangles CDE, CPA are similar (AAA), giving |DE|/|EC| = |PA|/|AC|.

Thus:

  |DE|.d =|EC||PA|     where d is the diameter

            = |EC||PE|     PA = PE since /AEP = 45 deg.

 

            = |BE||ET|     Intersecting chords

            = (|BD| – |DE|)(|BD| + |DE|)
            = |BD|^2 – |DE|^2
            = |FC|^2 – |DE|^2
            = |FG|(|FG| + d) – |DE|^2          Tangent/chord property

which can be rearranged to give:

    (|DE| – |FG|)(d + |FG| + |DE|) = 0

Since  d + |FG| + |DE| > 0, it follows that   |DE| = |FG|.


Edited on March 21, 2013, 3:39 pm

Edited on March 21, 2013, 3:44 pm
  Posted by Harry on 2013-03-21 15:36:41

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