Link AD, angle ADB = 90, PD=DE, APE is isosceles triangle, angle PAD=CAD, link CB, D is the midpoint of dazzling AC, angle CBP=DBA, angle PAD=DBA.

Angle CAB+CAD+DBA=90=PAD+DAC+CAB=90.

Second, if the perpendicular of AB passes through H through E, let EB=DE=PD=x, the triangle ADB is similar to PAB, and AB = times the square root of X,

EBH is similar to ABP, and it can be obtained that HB=3 times the square root x of 6. Therefore, under the root sign, AH=AB-HB, eh = EB 2-HB 2, and finally find tanCAB, and the angle CAB=BDC, it can be obtained. I didn't think much about this problem, and the solution is a bit troublesome. There should be a simpler solution.

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