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prove that am is perpendicular to bd

D=180 cause it a right angle

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.