(In reply to
Typo in problem statement by Bractals)
It's still pretty easy.
I'll call the distance sought h.
Area of triangle ABC = .5*AB*AD*sin(45º)
also
Area of triangle ABC = .5*AC*h
So .5*AB*AD*sin(45º) = .5*AC*h
h = √(.5)*AB*AD/AC
By law of cosines
AC²=AB² + BC²  2*AB*BC*cos(135º)
If you substitute this for AC as well as AD for BC the final expression for h is rather messy:
h = √(.5)*AB*AD/√(AB² + AD² + √(2)*AB*AD)
(Note: h can probably be written in terms of just AB, since the triangle ABD is uniquely determined by the ASA congruence theorem.)

Posted by Jer
on 20130516 00:38:14 