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|.
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
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Posted by Harry
on 2013-03-21 15:36:41 |
Different approach
