想半天没想出来

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想半天没想出来想半天没想出来 想半天没想出来(1)在ABE中,∠AED=1/2∠BAC+∠B在四边形EAGC中∠AED+∠EAD+G+1/2(∠B+∠BAC)+∠C=360°(四边形内角和)

想半天没想出来
想半天没想出来
 

想半天没想出来
(1)在ABE中,∠AED=1/2∠BAC+∠B
在四边形EAGC中
∠AED+∠EAD+G+1/2(∠B+∠BAC)+∠C=360°(四边形内角和)
(1/2∠BAC+∠B)+90°+(60°+∠DAE)+1/2(∠B+∠BAC)+∠C=360°  (只是代替)
 2×1/2∠BAC+∠B  +∠C++90°+(60°+∠DAE)+1/2∠B=360°
(前三项是三角形的内角和)
180+150+∠DAE+半∠B=360°
∴∠DAE+半∠B=30°
即∠DAE=30°-半∠B
(2)∵∠AED=90-∠DAE,        ∠AED=半∠BAC+∠B
∴90-∠DAE=半∠BAC+∠B  然后代入(1)的结果
  90°-30°+半∠B=半∠BAC+∠B    
∴半∠B+半∠BAC=60°
∴∠C=60°