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Junior high school mathematics right angle
Let d be DN perpendicular to AC, AC in n, AD and BM in p.

Angle APB= = Angle BAM = 90 degrees.

Angle ABM male * * *

So angle BAP= = angle BMA, angle MAD= = angle ABM.

Angle sum = = angle bam = 90 degrees.

So triangle ABM is similar to triangle NAD.

BA/AM=AN/DN=2=AN/NC

Because dn//ab

So an/NC = BD/DC = 2 = AB/MC.

Because angle ABD= = angle MCD = 45 degrees

So triangle ABD is similar to triangle MCD.

So angle BAD= = angle CMD= = angle AMB.

So angle AMB= = angle CMD