Let "a" denote the angle <BAH.
<HAC = a
AM is a bisector of "a" therefore <HAM=<MAD=a/2
Realize that triangle CDH is similar to triangle ABC and so <CHD=a
Now, run a line from B perpendicular to AC. Let K be the intersection of this line with AC.
Realize that triangle BCK is also similar to triangle ABC and so <CBK=a
Because |CB|=2|CH|, it holds that |CK|=2|CD| and so line BD is a bisector of <CBK and so <CBD=a/2
Let N denote the intersection of AM with BD. We now need to show that <AND=90 degrees. This is easy given the deduction above:
<ADB = 90 - (180 - a/2 - 180 + a) = 90 - a/2
<AND = 180 - <NAD - <ADB = 180 - a/2 - 90 + a/2 = 90 degrees => AM perpendicular to BD. This completes the proof.
