Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

m(AED )= 90°

Let ABC be an isosceles triangle with AB = AC. Let X ad Y be points on sides
BC and CA such that XY // AB. Denote by D the circumcenter of triangle
CXY and by E be the midpoint of BY . Prove that m(AED )= 90°.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: m(AED )= 90°

I got it.
Let the points be represented by numbers in a complex plane (a,b,c,d,x,y and s instead of E), and insert the coordinate axes such that D=0, the circumcircle of CXY is the unit circle and BC is parallell with the real axis. Let

,
|AB|=|AC|=R. The fact that XY||AB implies that XYC also is an isosceles triangle with |XY|=|CY|. From that we can easily conclude that y=i. We also have that:

so

we have that, since BC is parallell to the real axis:

Also, we have that

which yields:

We also have (for the complex number s representing the point E):

Now consider a-s, which represents the vector EA starting from the origin. If we rotate it pi/2 radians counter clockwise, and if AED=pi/2, the two vectors should be paralell. Rotation by pi/2 is just multiplication by i:

Now if this was parallell with DE, there should be a real (positive) constant k such that v=ks.  ie Re(v)=kRe(s), Im(v)=kIm(s). so starting with the imaginary parts:

Note also that k is positive. Now we just need to check that the equation for the real parts hold for this k. Indeed:

Thus

Q.E.D

Last edited by Kurre (2009-03-03 03:05:36)