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Mathematical problem-solving ability demonstration 20 15
Take EG=EB on EC and connect FG. de = ef,∴⊿GEF≌⊿BED,GF=BD,∠FGE=∠DBE。

∵gf=bd,bd=cf,∴gf=cf,∠c=∠fgc;

∵∠FGE=∠DBE,∴ Their complementary angle ∠FGC=∠ABC, then ∠ C = ∠ ABC is an isosceles triangle.

∫≈a = 58 ,∴∠c=( 180-58)÷2 = 6 1 .