三角形,ABC中,D是BC中点,任作一直线交AB,AD,AC,分别于P.N.Q求证AB/AP,AD/AN,AC/AQ,成等差数列
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![三角形,ABC中,D是BC中点,任作一直线交AB,AD,AC,分别于P.N.Q求证AB/AP,AD/AN,AC/AQ,成等差数列](/uploads/image/z/14334151-31-1.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2%2CABC%E4%B8%AD%2CD%E6%98%AFBC%E4%B8%AD%E7%82%B9%2C%E4%BB%BB%E4%BD%9C%E4%B8%80%E7%9B%B4%E7%BA%BF%E4%BA%A4AB%2CAD%2CAC%2C%E5%88%86%E5%88%AB%E4%BA%8EP.N.Q%E6%B1%82%E8%AF%81AB%2FAP%2CAD%2FAN%2CAC%2FAQ%2C%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
三角形,ABC中,D是BC中点,任作一直线交AB,AD,AC,分别于P.N.Q求证AB/AP,AD/AN,AC/AQ,成等差数列
三角形,ABC中,D是BC中点,任作一直线交AB,AD,AC,分别于P.N.Q求证AB/AP,AD/AN,AC/AQ,成等差数列
三角形,ABC中,D是BC中点,任作一直线交AB,AD,AC,分别于P.N.Q求证AB/AP,AD/AN,AC/AQ,成等差数列
过点B作BE‖PQ,交AD的延长线于点E;过点C作CF‖PQ,交直线AD于点F.
则有:AB/AP = AE/AN ,AC/AQ = AF/AN ,BE‖CF .
因为,BE‖CF ,DB = DC ,
所以,DE = DF .
因为,AB/AP + AC/AQ = AE/AN + AF/AN = (AD+DE)/AN + (AD-DF)/AN = 2AD/AN ,
所以,AB/AP,AD/AN,AC/AQ,成等差数列.
证明: 若要证明AB/AP,AD/AN,AC/AQ,成等差数列,只需证明 AD/AN - AB/AP = AC/AQ - AD/AN 即可。 设:①= AD/AN - AB/AP;②= AC/AQ - AD/AN 利用△的面积=1/2 *ab*sinC公式,可以证得: ①=(AD*AP - AB*AN)/ AN*AP = (S△ADP - S△ABN)/ S△ANP =(S△ABD - S△BDP - S△ABD + S△BDN)/ S△APN =(S△BDN - S△BDP)/ S△APN = BD*h2 / 2 * S△APN 同理可得:②=CD*h1 / 2 * S△ANQ ∴①:②= h2/h1 * S△ANQ/S△APN 如图所示: S△ANQ/S△APN=NQ:PN ∵△PNE∽△NQF ∴NQ:PN = h1:h2 ∴①:② = 1 ∴① = ② ∴AB/AP,AD/AN,AC/AQ,成等差数列