Ta co [TEX]\triangle ABC[/TEX] vuong tai A, [TEX]\widehat{ACB}=15^o[/TEX] (gt)
\Rightarrow [TEX]\widehat{ABC}=75^o[/TEX] (dli tam giac vuong)
O mien trong cua [TEX]\triangle OBC[/TEX] dung tam giac MBC deu
\Rightarrow [TEX]\widehat{MBC}=60^o[/TEX] (t/c tam giac deu)
Lay D la trung diem cua canh OB
\Rightarrow BD=OD=AC=[TEX]\frac{OB}{2}[/TEX]
Xet [TEX]\triangle DMB[/TEX] va [TEX]\triangle ABC[/TEX]
co BD=AC (c/m tren)
[TEX]\widehat{DBM}=\widehat{ACB}=15^o[/TEX]
BM=BC ([TEX]\triangle MBC[/TEX] deu theo cach dung)
nen [TEX]\triangle DBM=\triangle ABC[/TEX] (c.g.c)
\Rightarrow [TEX]\widehat{BDM}=\widehat{ABC}[/TEX] va [TEX]\widehat{DMB}=\widehat{ABC}[/TEX] (cac goc tuong ung)
do do [TEX]\widehat{BDM}=90^o, \widehat{DMB}=75^o[/TEX]
Ta c/m duoc [TEX]\triangle DBM=\triangle DOM[/TEX] (c.g.c)
\Rightarrow [TEX]\widehat{DMB}=\widehat{DMO}=75^o[/TEX] (2 goc tuong ung)
nen [TEX]\widehat{OMB}=75^o+75^o=150^o[/TEX]
ma [TEX]\widehat{OMB}+\widehat{BMC}+\widehat{OMC}=360^o[/TEX]
thay vao ta duoc
[TEX]150^o+60^o+\widehat{OMC}=360^o[/TEX]
\Rightarrow [TEX]\widehat{OMC}=150^o[/TEX]
Xet [TEX]\triangle OMB[/TEX] va [TEX]\triangle OMC[/TEX]
co OM la canh chung
[TEX]\widehat{OMB}=\widehat{OMC}=150^o[/TEX]
MB=MC (tam giac MBC deu)
nen [TEX]\triangle OMB=\triangle OMC[/TEX] (c.g.c)
\Rightarrow OB=OC (2canh tuong ung)
Vay [TEX]\triangle OBC [/TEX]can tai O (dinh nghia tam giac can)
@-)
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