如图,D是等边△ABC外一点,AD与BC交于点O,若角BDC=120°,BD=根号3-根号2,CD=根号3+根号2,则AO=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 06:00:08
![如图,D是等边△ABC外一点,AD与BC交于点O,若角BDC=120°,BD=根号3-根号2,CD=根号3+根号2,则AO=](/uploads/image/z/3710857-49-7.jpg?t=%E5%A6%82%E5%9B%BE%2CD%E6%98%AF%E7%AD%89%E8%BE%B9%E2%96%B3ABC%E5%A4%96%E4%B8%80%E7%82%B9%2CAD%E4%B8%8EBC%E4%BA%A4%E4%BA%8E%E7%82%B9O%2C%E8%8B%A5%E8%A7%92BDC%3D120%C2%B0%2CBD%3D%E6%A0%B9%E5%8F%B73-%E6%A0%B9%E5%8F%B72%2CCD%3D%E6%A0%B9%E5%8F%B73%2B%E6%A0%B9%E5%8F%B72%2C%E5%88%99AO%3D)
xS]OA+v3?liUhe1&>dAQcB$E?e[ʓ;-̝9dJ=ӽI?:`>/vӕDy~m@>k~'et*7N皨3W?f鷢exIE*,=TptŻUmg/5-?69VSRà
f8ebdk +VL(5(j5Xk2AErQu}lM4W
kp&..kaҹ8}B
.hϒr5ϒҸ":Ȟ|)De@aA(taZIB[w{z*YtsysluvAK>Уzlɠ6g111JZԲ@Z$؉C$P!Nذp
۔ըAd#
ItJ|%/VEy/H7vА1H)m30v^Wq,Ω/
如图,D是等边△ABC外一点,AD与BC交于点O,若角BDC=120°,BD=根号3-根号2,CD=根号3+根号2,则AO=
如图,D是等边△ABC外一点,AD与BC交于点O,若角BDC=120°,BD=根号3-根号2,CD=根号3+根号2,则AO=
如图,D是等边△ABC外一点,AD与BC交于点O,若角BDC=120°,BD=根号3-根号2,CD=根号3+根号2,则AO=
∵四边形ABDC中,∠BAC+∠BDC=180°,∴ABDC内接于圆,
延长DC到E,使CE=BD,连接AE,则∠ACE=∠ABD,又AC=AB,∴⊿ACE≌⊿ABD,
得AE=AD,∠4=∠ABC=60°,∴⊿ADE是等边三角形,
∴AD=DE=CD+CE=CD+BD=6..
注意到∠1=∠2;∠3=∠ACB=60°=∠4,得⊿ABD∽⊿COD, AD/CD=BD/OD.,
∴OD=BD*CD/AD=(3-√2)(3+√2)/6=7/6,
那么AO=AD-OD=6-7/6=29/6=4又5/6..