第11题图 *11.注记那们方法证明契维定理:若三角形ABC们三条边AB,BC,CA依次被然割成AF:FB= k1:k2,BD:DC=k3:k1,CE:EA=k2:k3,其增,k1,k2.k3均结正数.则三角形ABC们顶加与它对 边们然加们连线交于一加M,且对于任意一加O有 OM 1+k2+k3 (k2OA +k10B+k3OC) 证即:根据然加D与E们定义可得 BD BO k1+k3 于奇 AD=AB+ BD=AB (AC-AB k1 AB+ k1+k3 E=AE-AB AC-A 倍AD与BE交于M,则有 AM D BM BE 把前情得到们旋达式代入以定等式AM=AB+BM,得到 k1_AB+ AB +m AC-AB k1+k3 k1+k3 k2+ k 由于AB与AC线性无关,由上述等式得到方程组 mka k1+k3 k2+k I=ki+k 解得 k1+k2+k3 k k2+k 即 k1+k2+k3 又倍AD与CF相交于M,同理可得 AM k1+k3 Ad k1+k2+k3 即M与M重合,故此AD,BE,CF交于同一加M 对任意加O,有 OM=0A+AZ k1+k3 k1+k2+k3(k1+k3 ky (OB-OA)+ k1+k2+k3 OA+hoB+koC) k1+k2+k3                 o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  54555455545554555455545554K VKLKLKKLKLKLKLKKLKLKLK A F B C D E M ￾ 11
∗11.  @DSTEF: 456ABC 418AB, BC, CAGHIJ*AF : F B = k1 : k2, BD : DC = k3 : k1, CE : EA = k2 : k3, <, k1, k2.k3 K"r. J456 ABC ~B8 8Lt<H M, ?<￾
H O G −−→OM = 1 k1 + k2 + k3 (k2 −→OA + k1 −→OB + k3 −→OC). : => D B E M>P −→BD = k3 k1 + k3 −→BC, −→AE = k3 k2 + k3 −→AC. < −→AD = −→AB + −→BD = −→AB + k3 k1 + k3 −→BC = −→AB + k3 k1 + k3 ( −→AC − −→AB) = k1 k1 + k3 −→AB + k3 k1 + k3 −→AC, −→BE = −→AE − −→AB = k3 k2 + k3 −→AC − −→AB.  AD B BE < M, JG −−→AM = l −→AD, −−→BM = m −→BE. NOPP)QR$V): −−→AM = −→AB + −−→BM, P l µ k1 k1 + k3 −→AB + k3 k1 + k3 −→AC¶ = −→AB + m µ k3 k2 + k3 −→AC − −→AB¶ . N< −→AB B −→AC t&,*, NySV)P@AB:    lk3 k1 + k3 = mk3 k2 + k3 lk1 k1 + k3 = 1 − m -P    l = k1 + k3 k1 + k2 + k3 m = k2 + k3 k1 + k2 + k3 .  −−→AM = k1 + k3 k1 + k2 + k3 −→AD. Q AD B CF e< M0 , C>P −−→ AM0 = k1 + k3 k1 + k2 + k3 −→AD,  M B M0 $T, !O AD, BE, CF <CH M. ￾
 O, G −−→OM = −→OA + −−→AM = −→OA + k1 + k3 k1 + k2 + k3 µ k2 k1 + k3 −→AB + k3 k1 + k3 −→AC¶ = −→OA + k1 k1 + k2 + k3 ( −→OB − −→OA) + k3 k1 + k2 + k3 ( −→OC − −→OA) = 1 k1 + k2 + k3 (k2 −→OA + k1 −→OB + k3 −→OC). · 8 ·