a, Xet tg vuong HAB va tg vuong KAC co:
AB=AC(hai canh ben tg can ABC)
BAC chung
Do do tg vuong HAB=tg vuong HAC(ch-gn)
b,Vi tg HAB=tg KAC (cau a)
=>HA=KA(2 canh t/u)
Xet tg AKH co :
HA=KA(cmt)
=>tg HAK can tai A(d/n tg can)
=>AKH= \(\dfrac{180^0-BAC}{2}\) (1)
Vi tg ABC can tai A :
=> ABC= \(\dfrac{180^0-BAC}{2}\) (2)
Tu (1) va (2)
=> AKH =ABC mà 2 góc này ở vị trí hai góc đồng vị
=>KH//BC(dpcm)
c, Xet tg vuong AKM va tg vuong AMH co:
AM chung
KA=AH(cau a)
do do :tg vuong KAM=tg vuong HAM(ch-gn)
=>KAM=HAM(2 goc t/u)
Ma tia AMnam giua hai tia KAva AH
=>AM la tia phan giac KAH (3)
Xet tg BAMva tg ACN co:
AB=AC(2 canh ben tg ABC can)
ABC=ACB(2 goc o day tg can ABC)
BN=CN(N la td BC)
Do do tg ABN=tg ACN(c.g.c)
=>BAN=NAC(2 goc t/u)
Ma tia AN nam giua hai tia AB va AC
=> AN la tia p/g BAC (4)
Tu (3) va (4)
=>3 diem A,M,N thang hang
