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 = (xy)/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 = (xy)/2
DI = (x+y)/2
Since HIE is a right triangle
EI = √((x+2)^2/2  (xy)^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 20130320 22:18:31 