Question

如图,已知线段AD,BC相交于点Q,DM平分∠ADC,BM平分∠ABC,且∠A=27°,∠M=33°,求∠C的度数

Answer 1

∵∠AQB=∠CQD

∴∠C+∠ADC=∠A+∠ABC

∠C=∠A+∠ABC-∠ADC

同样地,∠A+∠ABM=∠M+∠ADM

即 2∠A+∠ABC=2∠M+∠ADC

∠ABC-∠ADC=2∠M-2∠A

∴∠C=∠A+2∠M-2∠A=2∠M-∠A=2×33°-27°=39°