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|.

Call AD = x and CD = y
So AM = CM = GM = (x+y)/2
MD = (x-y)/2
BD = FC = √xy
Since MCF is a right triangle
MF = √(xy+((x+y)/2)^2)
= √(x^2+6xy+y^2)/2
FG = MF - GM = √(x^2+6xy+y^2)/2 - (x+y)/2

Now to get our 135 degree angle = an angle that inscribes 270 degrees of arc.

Construct point H on the circumcircle with HM perpendicular to AC.  Angle AHC is right.
Construct a circle with center H containing A (and C).
The major arc from A to C on this new circle is 270 degrees.
Point E is where this circle intersects BD, which guarantees angle AEC is 270/2 = 135 degrees.

Extend the altitude BD to a line.  Construct point I that makes DMHI a rectangle. 
HE = √2 * (x+y)/2
HI = (x-y)/2
DI = (x+y)/2
Since HIE is a right triangle
EI = √((x+2)^2/2 - (x-y)^2/4)
=√(x^2+6xy+y^2)/2

Finally DE = EI - DI = =√(x^2+6xy+y^2)/2 - (x+y)/2
Which is exactly the same as FG above.

QED

Posted by Jer on 2013-03-20 22:18:31