Certificate: It is known that in △ABC, ∠ C = 90, point M is on BC, point BM=AC, point N is on AC, point AN=MC, and point AM and BN intersect.

Let AC=BM=X and MC=AN=Y, then

BC=BM+MC=X+Y，CN=AC-AN=X-Y

AM=√(AC^2+MC^2)=√(X^2+Y^2)

If the N-crossing point is NE⊥AM and the E-crossing point is AM, then △AEN∽△ACB.

AE/AN=AC/AM，NE/AN=MC/AM

AE=AN*AC/AM=Y*X/√(X^2+Y^2)

NE=AN*MC/AM=Y^2/√(X^2+Y^2)

If the intersection p is PF⊥BC and the intersection BC is f, then △PFM∽△ACM, △BPF∽△BNC.

PF/FM=AC/MC，PF=FM*AC/MC=FM*X/Y

PF/BF=CN/BC，PF=BF*CN/BC=BF*(X-Y)/(X+Y)

BF*(X-Y)/(X+Y)=FM*X/Y

BF =(FM * X/Y)*[(X+Y)/(X-Y)]= FM * X *(X+Y)/[Y *(X-Y)]

BF=BM+FM=X+FM

FM*X*(X+Y)/[Y*(X-Y)]=X+FM

FM=XY*(X-Y)/(X^2+Y^2)

PM/FM=AM/CM

pm=fm*am/mc=[xy*(x-y)/(x^2+y^2)]*[√(x^2+y^2)/y]

=X*(X-Y)/√(X^2+Y^2)

PE=AM-AE-PM

=√(x^2+y^2)-y*x/√(x^2+y^2)-x*(x-y)/√(x^2+y^2)

=Y^2/√(X^2+Y^2)

= Northeast

Because NE⊥AM, which is NE⊥PE.

It is known that in the right angle △NEP, NE=PE.

Therefore ∠ EPN = 45 degrees.

But ∠BPM=∠EPN.

So BPM = 45.

Evidence 2:

Certificate: It is known that in △ABC, ∠ C = 90, point M is on BC, point BM=AC, point N is on AC, point AN=MC, and point AM and BN intersect.

Let AC=BM=X and MC=AN=Y, then

BC=BM+MC=X+Y，CN=AC-AN=X-Y

tan∠AMC=AC/MC=X/Y

tan∠NBC=CN/BC=(X-Y)/(X+Y)

∠AMC=∠BPM+∠NBC

∠BPM=∠AMC-∠NBC

tan∠BPM=tan(∠AMC-∠NBC)

=(tan∠AMC-tan∠NBC)/( 1+tan∠AMC * tan∠NBC)

=[X/Y-(X-Y)/(X+Y)]/[ 1+(X/Y)*(X-Y)/(X+Y)]

=[X *(X+Y)-Y *(X-Y)]/[Y *(X+Y)+X *(X-Y)]

=(X ^2+Y ^2)/(X ^2+Y ^2)

= 1

Because BPM

So BPM = 45.