Question

如圖13,已知AB為半圓O的直徑,AP為過(guò)點(diǎn)A的半圓的切線,在上任取一點(diǎn)C(點(diǎn)C與A、B不重合),過(guò)點(diǎn)C作半圓的切線CD交AP于點(diǎn)D;過(guò)點(diǎn)C作CE⊥AB,垂足為E,連結(jié)BD,交CE于點(diǎn)F.

         

(1)　　                   (2)

圖13

(1)當(dāng)點(diǎn)C為的中點(diǎn)時(shí)(如圖13(1)),求證:CF =EF;

(2)當(dāng)點(diǎn)C不是的中點(diǎn)時(shí)(如圖13(2)),試判斷CF與EF的相等關(guān)系是否保持不變,并證明你的結(jié)論.

Answer 1

思路分析:第(1)題E與O重合,只需證明四邊形DAEC為矩形,CD∥AB即可.?

(2)由(1)的結(jié)論猜測(cè)CF =EF仍然成立.然后再設(shè)法證明.

證明:(1)∵DA是切線,AB為直徑,∴DA⊥AB.?

∵點(diǎn)C是的中點(diǎn),且CE⊥AB,?

∴CE過(guò)圓心.∴點(diǎn)E為半圓的圓心.?

又∵DC是切線,∴DC⊥EC.?

∴四邊形DAEC為矩形.?

∴CE∥AD且CE =AD.?

∴= =,即 =,?

∴F為EC的中點(diǎn),即CF =EF.?

(2)CF =EF仍然成立.證明如下:?

連結(jié)BC并延長(zhǎng)交AP于G點(diǎn),連結(jié)AC.?

∵AD、CD是半圓的切線,?

∴DC=DA.∴∠DAC=∠DCA.?

∵AB為直徑,?

∴∠ACB =90°,∠ACG =90°,∠G+∠DAC =∠DCA +∠DCG =90°.?

∴∠G = ∠DCG.?

在△GDC中,GD =DC,又∵DC =DA,∴GD =DA.?

∵AP是半圓O的切線,?

∴AP⊥AB.又CE⊥AB,∴CE∥AP.?

∴= =.?

又GD =AD,∴CF =EF.