Question

    問題提出

如圖①，已知直線l與線段AB平行，試只用直尺作出AB的中點．

初步探索

如圖②，在直線l的上方取一個點E，連接EA、EB，分別與l交于點M、N，連接MB、NA，交于點D，再連接ED并延長交AB于點C，則C就是線段AB 的中點．

推理驗證

利用圖形相似的知識，我們可以推理驗證AC＝CB．

（1）若線段a、b、c、d長度均不為0，則由下列比例式中，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

如圖③，△ABC中，D是BC的中點，點P在AB上．

（3）在圖③中只用直尺作直線l∥BC．

（4）求證：l∥BC．

 

Answer 1

解：（1）B；········································································································· 1分

（2），；·························································································· 3分

（3）如圖①，畫圖正確；················································································· 6分

（4）如圖②，過點Q作MN∥BC，交AB、AC分別于點M、N，

∵MN∥BC，

∴△AMQ∽△ABD，△AQN∽△ADC，

∴＝ ，＝，

∴＝ ．···················································································· 7分

∵點D是BC的中點，

∴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