初一数学暑假作业解答题.如图，若
(1)做直线BE,使BE//AM//CN,(E在B的右边)则有∠MAB+∠ABC+∠BCN= ∠MAB+(∠ABE+∠CBE)+∠BCN=( ∠MAB+∠ABE)+(∠CBE+∠BCN)=180°+180°=360°
(2)做直线A1 M1,A2M2,使A1 M1//A2M2//A1M//CN,
则有∠MAA1+∠AA1A2+∠A1A2C+∠A2CN
=∠MAA1+(∠AA1M1+∠M1A1A2)+(∠A1A2M2+∠M2A2C)+∠A2CN
=(∠MAA1+∠AA1M1)+(∠M1A1A2+∠A1A2M2)+(∠M2A2C+∠A2CN)
=180°+180°+180°=540°
(3)可...全部
(1)做直线BE,使BE//AM//CN,(E在B的右边)则有∠MAB+∠ABC+∠BCN= ∠MAB+(∠ABE+∠CBE)+∠BCN=( ∠MAB+∠ABE)+(∠CBE+∠BCN)=180°+180°=360°
(2)做直线A1 M1,A2M2,使A1 M1//A2M2//A1M//CN,
则有∠MAA1+∠AA1A2+∠A1A2C+∠A2CN
=∠MAA1+(∠AA1M1+∠M1A1A2)+(∠A1A2M2+∠M2A2C)+∠A2CN
=(∠MAA1+∠AA1M1)+(∠M1A1A2+∠A1A2M2)+(∠M2A2C+∠A2CN)
=180°+180°+180°=540°
(3)可以。
两条平行线之间有N个点,则点A,点C与这N个点依次连线所形成的N+2个角的和为180°×(N+1)
如第一题,两条平行线之间有一个点B,依次连接AB,BC,所形成的三个角(N=1,N+2=3,所以是三个角)的和为180°×(N+1)=180°×(1+1)=360°
如第二题,两条平行线之间有两个点A1,A2,依次连接AA1,A1A2,A2C,所形成的四个角(N=2,N+2=3,所以是四个角)的和为180°×(N+1)=180°×(2+1)=540°
。
收起
