Let ABCD be a parallelogram with ∠BAC = 45° and ∠ABD = 30°.

What is the distance from B to diagonal AC in terms of the
lengths of sides AB and AD?

(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 2013-05-16 00:38:14