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Mathematical problems (difficulties)
Please look at the picture to do triangular BPM? And then what? High line peak of APN? PF。

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.