解必程与 ∫-1-m = 得k=m=景·所以G的坐不为（信，）同解可以推得G的坐不为（传）证得G=G. 第4题图 第5题图 5.证明三角形的三条角平分线证于一点. 证明设△ABC的三条角平分线分别为AD,BE和CF.且设AD与BE证于T点.令 示=k而网+G+G, 电立物时不等认成且令=，衣-了则7坐不(T我还 知道 配-+远风+6. 1 所以 B所=mB正= 1+16--6-+石- m =-md+ 由于7=丽+立，所以 多 1a+1-1-m m a (++1-可 解得 k= 1a1+61 @1+11+1元- 又设AD与CF证于T'点， A7=$而= @+1+ s 得T点的坐不为+1+可】 CT=CF= 石 +面+0=6+-- 10 -@AB ( k 2 = 1 − m k 2 = m 2 P k = m = 2 3 . #$ G WU" ³ 1 3 , 1 3 ´ . C, >$^P G0 WU" ³ 1 3 , 1 3 ´ . SP G = G0 .       o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  666666666666 V 6 M 66666676666666 NMMMNMMMNMMMNMMMNMM A F B C DE M ￾ 4
                
o n o o n o o n o o n o o n o o n (uuuuuuuu ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( D  565565565565565565565565565 %YSSSSSSSSSSSSSSSSS A F B C D E M ￾ 5
5. ST456415t<H. :  4ABC 415t" AD, BE : CF. ? AD B BE < T . I −→AT = k −→AD = k | −→AB| + | −→AC| (| −→AC| −→AB + | −→AB| −→AC). [+\]UV [A; −→AB, −→AC], ?I −→AB = −→a , −→AC = −→b . J T WU Ã k| −→b | | −→a | + | −→b | , k| −→a | | −→a | + | −→b | ! . _ ` −→BE = 1 | −→BA| + | −→BC| (| −→BC| −→BA + | −→BA| −→BC), #$ −→BT = m −→BE = m | −→a | + | −→b − −→a | (−|−→b − −→a | −→a + | −→a |( −→b − −→a )) = −m−→a + m| −→a | | −→a | + | −→b − −→a | −→b . N< −→AT = −→AB + −→BT, #$ Ã k| −→b | | −→a | + | −→b | , k| −→a | | −→a | + | −→b | ! = (1, 0) + Ã −m, m| −→a | | −→a | + | −→b − −→a | ! , :    k| −→b | | −→a | + | −→b | = 1 − m k| −→a | | −→a | + | −→b | = m| −→a | | −→a | + | −→b − −→a | -P: k = | −→a | + | −→b | | −→a | + | −→b | + | −→b − −→a | . Q AD B CF < T 0 , −−→ AT0 = s −→AD = s| −→b | | −→a | + | −→b | −→a + s| −→a | | −→a | + | −→b | −→b . P T 0 WU" Ã s| −→b | | −→a | + | −→b | , s| −→a | | −→a | + | −→b | ! . −−→ CT0 = t −→CF = t | −→CA| + | −→CB| (| −→CB| −→CA + | −→CA| −→CB) = t| −→b | | −→b | + | −→a − −→b | −→a − t −→b , · 10 ·