Why not come to FC=PE? Because they are parallelograms.
Suppose PE=NC? Then f and c must be at the same point.
BM*PE/2=2*( 1/2*BM)*PF (multiplied by a 0. The reason of 5 is that their ratio of height to bottom is 1:2? So the area is naturally multiplied by 2)
{ BM *( 1/2 * BM)}/2 = 2 *( 1/2 * BM)* PF
BM=2PF
So PF=PE? (because BM=2*NC=2*PE)
And because PF=AN
Is triangle APN an isosceles right triangle? So APN angle =45 degrees.
Prove it here Did you get the angle BPM=45 degrees? But what about premise F? Will it be at the same point as n?
The answer is no?
The reason is that angle c equals 90 degrees? What if n and f are at the same point? So PFC equals 90? Isn't PNC 90?
So triangle ABC is not an isosceles right triangle? ? And this isosceles triangle, his waist must be AB and BC? Because there can be a line as the center line? Only the midline/vertical line of isosceles triangle from top to bottom can be used as the vertical line.
So the base angle of this isosceles triangle is angle C=90? 2*90= 180? The sum of two angles can be equal to 90 degrees? So it can't be a triangle?
So the answer angle BPM is less than 45.