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Mathematical problems of triangle in the second day of junior high school
Proof: connect AM.

△ ABC is an isosceles right triangle, and m is the midpoint of the hypotenuse BC.

∴am⊥bc,am=bm,△abm≌△cam,∠mac=∠mba=45

∠∠b = 45,DF⊥BF.

∴△BFD is an isosceles right triangle.

∵DF⊥AB,DE⊥AC,∠BAC=90?

A quadrilateral is a rectangle.

∴ BF = FD = AE and ∠ FBD = 45.

∫△BFM is equal to△△△ AEM (SAS).

∴∠BMF=∠AME,FM=EM

∴∠FME=∠FMA+∠AME=∠BMF+∠FMA=90

△ FME is an isosceles right triangle.