Bài 6:
\(\widehat{ABC}=180^0-90^0-30^0=60^0\)
nên \(\widehat{DBC}=40^0\)
\(\Leftrightarrow\widehat{BDC}=180^0-40^0-30^0=110^0\)
Xét ΔDBC có \(\widehat{C}< \widehat{DBC}< \widehat{BDC}\)
nên BD<DC<BC
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