Question

初二数学题!有过程!先回答!有好评!急急急

Answer 1

∵∠C=90°
∴∠CAB+∠CBA=90°
∴∠EAB+∠FBE=360°-∠CAB-∠CBA
=360°-90°
=270°
又∵AD、BD分别平分∠EAB、∠FBE
∴∠DAB+∠DBA=270°/2
=135°
∵∠D=180°-∠DAB-∠DBA
=45°

Answer 2

角EAB=90+角B
角ABF=90+角A
角EAB+角ABF=180+角A+角B=270
角D=180-（270/2）=45度。

Answer 3

由题意得两角平分线交于点D,设角CAB为X,角ACB为Y,角D为Z.由图可列方程①（90°+X）÷2+（90°+Y）÷2+Z=180° ②90°+X+Y=180° 由①得180°+X+Y+2Z=360° 解得X+Y+2Z=180°③ 将③带入②中得90°+2Z=180°解得Z=45° 所以角D等于45°