Question

已知：△ABC中，∠B、∠C的角平分线相交于点D，过D作EF//BC交AB于点E，交AC于点F．求证：BE+CF=EF

详细回答

Answer 1

∵BD是∠B的角平分线，∴∠EBD=∠CBD

同理，∵CD是∠C的角平分线 ∴∠FCD=∠BCD

又∵EF‖BC→∠BDE=∠CBD,即∠BDE=∠EBD→△BDE中BE=DE...①, 

            ∠CDF=∠BCD,即∠CDF=∠FCD→△CDF中CF=DF...②.

  由①②以及EF=DE+DF,可得BE+CF=DE+DF=EF