题目内容
问题提出
如图①，已知直线l与线段AB平行，试只用直尺作出AB的中点．
初步探索
如图②，在直线l的上方取一个点E，连接EA、EB，分别与l交于点M、N，连接MB、NA，交于点D，再连接ED并延长交AB于点C，则C就是线段AB 的中点．
推理验证
利用图形相似的知识，我们可以推理验证AC＝CB．
（1）若线段a、b、c、d长度均不为0，则由下列比例式中，一定可以得出b＝d的是（）
| A． | B． | C． | D． |
＝
＝
（2）由MN∥AB，可以推出△EFN∽△ECB，△EMN∽△EAB，△MND∽△BAD，
△FND∽△CAD．
所以，有
＝
＝
＝＝
，
所以，AC＝CB．
拓展研究
如图③，△ABC中，D是BC的中点，点P在AB上．
（3）在图③中只用直尺作直线l∥BC．
（4）求证：l∥BC．
 |
解：（1）B；········································································································· 1分
（2）
，
；·························································································· 3分
（3）如图①，画图正确；················································································· 6分
（4）如图②，过点Q作MN∥BC，交AB、AC分别于点M、N，
∵MN∥BC，
∴△AMQ∽△ABD，△AQN∽△ADC，
∴
＝
，＝
，
∴＝ ．···················································································· 7分
∵点D是BC的中点，
∴BD＝CD，
∴MQ＝NQ．
∵MN∥BC，
∴△PMQ∽△PBC，△EQN∽△EBC，
∴
＝
，
＝
， ∴＝，
∴
＝
，······················································································· 8分
又∵∠PQE＝∠CQB，
∴△PQE∽△CQB，················································································· 9分
∴∠EPQ＝∠BCQ，
∴PE∥BC，
即l∥BC．······························································································ 10分
 |
阅读快车系列答案
,将△BCD沿
方向平移,得到△EFG.
如图，是某卖场国产大米牌手机的宣传广告．
 |
（1）你认为大米手机5月份的销售量必定是三个品牌手机中最高的吗？通过计算说明你的理由．
（2）若各品牌手机2015年4月的销售量如下：
| 手机品牌 | 芒果手机 | 四星手机 | 大米手机 |
| 销售量（台） | 200 | 80 | 120 |
求该卖场5月份三个品牌手机销售量的平均增长率．
