1, let AH=AE be on AB, cross e and make parallel lines of AB intersect BC at g, then the triangle BHE congruent triangles EDF, conclusion: EB=EF.

(angle AHC=a, angle BHC= 180-a= angle d, BH=ED, angle ABE= angle BEG, angle BEG+ angle GEF= angle DEF+ angle GEF, congruence condition: angle).

2. Extend AB to H to make AH=AE and connect EH, then the triangle BHE similar triangles EDF, BE/EF=AH/DE=n-m+ 1.

This question is similar to the previous senior high school entrance examination in which city, but it is a right triangle. I've seen it. The solution to the problem is the same as before.

Description: Dalian Carnegie eloquence major is engaged in speech eloquence training. The lecturer is studying mathematics. He did this problem, but I can't. According to what he wrote, I typed it.