Question

如图,已知点BC分别在射线AN,AM上,∠MCB与∠NBC的平分线交于点P 求证

Answer 1

作PE⊥AM于E,PF⊥BC于F,PG⊥AN于G,

∠MCB与∠NBC的平分线交于点P,

∴PE=PF=PG,

∴AP平分∠BAC.

设AP=x,在△ABP中,∠BAP=45°,AB=7,PB=5,由余弦定理,

25=49+x^2-7√2x,

∴x^2-7√2x+24=0,

解得x1=4√2,x2=3√2(舍).

∴AP=4√2.