â³ABCï¼D.E.Fåå«æ¯AB BC CAçä¸ç¹ï¼BFä¸CD交ä¸ç¹O,设åéAB=a,AC=b,è¯æAOEä¸ç¹å
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ç±äºåé符å·ä¸å¥½åï¼ä»¥ABè®°ABåéï¼AB=-BA
设ä¸çº¿AEä¸BF交äºO
设AO=mAE=m(AC+CE)=(m/2)(2AC-BC)=(m/2)(2AC+CB)
设BO=nBF=n(BC+CE)=(n/2)(2BC-AC),OB=-BO=(n/2)(AC+2CB)
AB=AO+BO=(m+n/2)AC+(m/2+c)BC
åAB=AC+CB
å³(m+n/2)AC+(m/2+c)BC=AC+CB
(-1+m+n/2)AC+(-1+m/2+c)BC=0
å 为AC,BCä¸å
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æ以m+n/2=m/2+n=1, 解å¾m=n=2/3
AO=(2/3)AA1
å设AEä¸CD交äºO1,åçå¯å¾AO1=(2/3)AE
Oä¸O1éå
æ以AE,BF,CD交äºä¸ç¹
ä¸è§å½¢ä¸è¾¹ä¸çº¿äº¤äºä¸ç¹.
ä¹å°±æ¯AOE ä¸ç¹å
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