Question

已知,如图,AC平分∠DAB.CE⊥AB,AB+AD=2AE.求证:∠D+∠B=180°

Answer 1

作CF⊥AD的延长线于F,

由AC平分∠DAB,CE⊥AB,CF⊥AD

得ACE=ACF,故AE=AF

因AB+AD=2AE=AE+AF,得BE=DF

又因BE=DF,CE=CF,∠CEB=∠CFD=90°

得BCE=DCF,∠BCE=∠DCF

因∠ACE=90°,得:∠ACE+∠CAE=90°

同理:∠ACE+∠CAF=90°,故∠CAF+∠ACD+∠DCE =90°

故∠BAD+∠BCF=180°

则在ABCD中∠ADC+∠ABC=180°

命题得证!