a.Ta có:
ˆBID=12ˆBIC=12(180o−ˆBCI−ˆIBC)=12(180o−12ˆBCA−12ˆABC)=12(180o−12(ˆBCA+ˆABC)=12(180o−12(180o−ˆBAC)=60oBID^=12BIC^=12(180o−BCI^−IBC^)=12(180o−12BCA^−12ABC^)=12(180o−12(BCA^+ABC^)=12(180o−12(180o−BAC^)=60o
Lại có :
ˆNIB=ˆIBC+ˆICB=12ˆABC+12ˆACB=12(ˆABC+ˆACB=12(180o−ˆBAC)=60oNIB^=IBC^+ICB^=12ABC^+12ACB^=12(ABC^+ACB^=12(180o−BAC^)=60o
→ˆNIB=ˆBID→NIB^=BID^
→ΔNIB=ΔDIB(g.c.g)→ΔNIB=ΔDIB(g.c.g)
→BN=BD→BN=BD
b.Chứng minh tương tự câu a
→CD=CM→CD=CM
→BN+CM=BD+CD=BC→đpcm→BN+CM=BD+CD=BC→đpcm