As shown in figure a, CE / / AB, so ∠ 1 = 2 = B, so ∠ ACD = 1 + ∠ 2 = a + B, which is a useful conclusion, In the quadrilateral ABCD of figure B, do AE / / BC to intersect DC with e through A. if you have this conclusion, find the degree of ∠ a + ∠ B + ∠ C + ∠ D Figure a http://hiphotos.baidu.com/yalijudy/pic/item/099efed1bbb16a299a50274f.jpg Figure B http://hiphotos.baidu.com/yalijudy/pic/item/6750e5f54b044b34bd3109e2.jpg

As shown in figure a, CE / / AB, so ∠ 1 = 2 = B, so ∠ ACD = 1 + ∠ 2 = a + B, which is a useful conclusion, In the quadrilateral ABCD of figure B, do AE / / BC to intersect DC with e through A. if you have this conclusion, find the degree of ∠ a + ∠ B + ∠ C + ∠ D Figure a http://hiphotos.baidu.com/yalijudy/pic/item/099efed1bbb16a299a50274f.jpg Figure B http://hiphotos.baidu.com/yalijudy/pic/item/6750e5f54b044b34bd3109e2.jpg

1) An outer angle of a triangle is equal to the sum of two inner angles not adjacent to it
2)AE//BC
∠BAC+∠B=180
∠DAE+∠D=∠AEC
∠AEC+∠C=180
∠DAB+∠B+∠C+∠D
=∠DAE+∠BAE+∠B+∠C+∠D
=(∠DAE+∠D)+∠BAE+∠B+∠C
=(∠BAE+∠B)+(∠C+∠DEA)=360