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

What is the distance from B to diagonal AC in terms of just |AB|?

What is the distance from B to diagonal AC in terms of just |AD|?

It feels as if there should be a simple geometry proof here.
I can’t find one - so here’s some trig:

Denote the lengths of AB, AD by x and y, and the angle CAD by Z.
Let p be the length of the perpendicular from B to AC.

Using the Sine Rule in triangles ABC and ABD:

y/x = sin(45^o – Z)/sin Z  = sin 30^o/sin 105^o

Solving for Z:   sin Z = sin(45^o – Z) sin(45^o + 60^o)/sin 30^o

            2 sin Z = (cos Z – sin Z)(1 + sqrt3)

            tan Z = (1 + sqrt3) / (sqrt3 + 3)  = 1 / sqrt3

            Z  = 30^o

So                    p = y sin 30^o  =  |AD|/2

and                   p = x sin 15^o = |AB|sqrt2(sqrt3 - 1)/4 

Posted by Harry on 2013-06-14 16:34:52