Question

如图2-25,经过点A、C的圆O切AB于点A,交CB于点D,∠ABC的平分线BG交⊙O于点G,作∠GAE =∠GAB,AE交CG于F点,交CD于E点,求证:

图2-25

(1)∠EDG =∠EFG;

(2)AG² =CG·DG.

Answer 1

思路分析:(1)要证∠EDG =∠EFG,只需证∠GDB =∠AFG,而∠GDB =∠CAG,∠AFG =∠GCA +∠CAF,由∠GAB为弦切角,用弦切角定理证明即可;?

(2)由(1)易证△AFG∽△CAG,从而得AG²=CF·CG,故再证FG =DG,连结EG,证△EFG≌△EDG就能达到目的.

证明:(1)∵四边形AGDC是圆内接四边形,?

∴∠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)连结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的距离相等(即G为△ABE的内心).?

∴G在∠AEB的角平分线上.?

∴∠FEG =∠DEG.?

∵EG =EG,∠EDG =∠EFG,∴△FEG≌△DEG.?

∴FG =DG.?

∴=,即AG²=CG·DG.