Xét tứ giác ABCD có \(\widehat{BAD}+\widehat{ADC}+\widehat{ABC}+\widehat{BCD}=360^0\)
=>\(\widehat{ABC}+\widehat{BCD}+70^0+50^0=360^0\)
=>\(\widehat{ABC}+\widehat{BCD}=240^0\)
=>\(\widehat{MBC}+\widehat{MCB}=\dfrac{240^0}{2}=120^0\)
Xét ΔMBC có \(\widehat{MBC}+\widehat{MCB}+\widehat{BMC}=180^0\)
=>\(\widehat{BMC}+120^0=180^0\)
=>\(\widehat{BMC}=60^0\)
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