Ta có: \(\widehat{BAD}=\widehat{CAD}=\dfrac{\widehat{BAC}}{2}\)
\(\widehat{ACB}=\dfrac{\widehat{BAC}}{2}\)
Do đó: \(\widehat{DAC}=\widehat{DCA}\)
Xét ΔDAC có \(\widehat{ADB}\) là góc ngoài tại đỉnh D
nên \(\widehat{ADB}=\widehat{DAC}+\widehat{DCA}\)
=>\(2\cdot\widehat{ACB}=70^0\)
=>\(\widehat{ACB}=35^0\)
\(\widehat{BAC}=2\cdot35^0=70^0\)
Xét ΔBAC có \(\widehat{BAC}+\widehat{ACB}+\widehat{ABC}=180^0\)
=>\(\widehat{ABC}+35^0+70^0=180^0\)
=>\(\widehat{ABC}=75^0\)
Đúng 2
Bình luận (1)