Question

求解，在线等

Answer 1

角E=180°-角EAB-角EBA
=180°-角EAC-角CAB-1/2角CBA
=180°-角DAF-角CAB-1/2角CBA
=180°-1/2角DAB-角CAB-1/2角CBA
=180°-1/2(180°-角CAB) -角CAB-1/2角CBA
=90°+1/2角CAB-角CAB-1/2角CBA
=90°-1/2角CAB-1/2角CBA
=90°-1/2(角CAB+角CBA)
=90°-1/2(90°)
=45°

Answer 2

设∠ABC=a,∠ABE=a/2,则∠CAB=90°-a,∠DAB=180°-∠CAB=180°-(90°-a)=90°+a,那么∠BAF=0.5*∠DAB=45°+a/2,∠EAB=180°-∠BAF=180°-(45°+a/2)=135°-a/2,在三角形ABE中就知道两个内角了，那么∠E=180°-EAB-EBA=45°

Answer 3

E坑
AB=1坑2坑
角E=3坑