Question

如圖2-25,經(jīng)過點(diǎn)A、C的圓O切AB于點(diǎn)A,交CB于點(diǎn)D,∠ABC的平分線BG交⊙O于點(diǎn)G,作∠GAE =∠GAB,AE交CG于F點(diǎn),交CD于E點(diǎn),求證:

圖2-25

(1)∠EDG =∠EFG;

(2)AG² =CG·DG.

Answer 1

思路分析:(1)要證∠EDG =∠EFG,只需證∠GDB =∠AFG,而∠GDB =∠CAG,∠AFG =∠GCA +∠CAF,由∠GAB為弦切角,用弦切角定理證明即可;?

(2)由(1)易證△AFG∽△CAG,從而得AG²=CF·CG,故再證FG =DG,連結(jié)EG,證△EFG≌△EDG就能達(dá)到目的.

證明:(1)∵四邊形AGDC是圓內(nèi)接四邊形,?

∴∠EDG +∠CAG =180°.?

∵∠EFG =∠AFC,?

∴∠EFG +∠ACF +∠CAF =180°.?

∵∠ACF =∠BAG,∠BAG =∠FAG,?

∴∠EFG +∠BAG+∠CAF =∠EFG+∠FAG +∠CAF =180°.?

∴∠EFG +∠CAG =180°.∴∠EDG =∠EFG.?

(2)連結(jié)EG,?

∵∠AFG =∠ACF +∠CAF,∠ACF =∠BAG =∠FAG,?

∴∠AFG =∠CAG.?

又∵∠AGC公用,∴△AGF∽△CGA.?

∴=.?

∵GA平分∠EAB,BG平分∠ABD,?

∴G到AE、B、BE的距離相等(即G為△ABE的內(nèi)心).?

∴G在∠AEB的角平分線上.?

∴∠FEG =∠DEG.?

∵EG =EG,∠EDG =∠EFG,∴△FEG≌△DEG.?

∴FG =DG.?

∴=,即AG²=CG·DG.