解方程与 得k=m=3所以G的坐标为(,3)同解可以推得C的坐标为(3,3)证得G=C 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 AT=kAD k (AC AB+JABJAC 1AB|+|4 建你法,且令=、正则7坐(同下)数的是 知道 BA (BCIBA+BABC 4+|BC 所以 =mBE la1+ b-al (-|b-da+|a(b-a) -ma'+-ma16 la+b-a 由于AT=AB+BT,所以 k =(1,0)+-m d+|b1|a+b1 la+ b-a 明 klbl la+b a+b-a 解得: Ial+b|+b-a 又设AD与CF证于T点, r=s万=-4b17+_7Lb la+b 得点的坐标为( +|b1’|+|b1 可面师麻+ 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 ·