Question

Answer 1

a: Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(180^0-3\cdot\widehat{C}+\widehat{C}+70^0=180^0\)

=>\(70^0-2\cdot\widehat{C}=0\)

=>\(2\cdot\widehat{C}=70^0\)

=>\(\widehat{C}=\dfrac{70^0}{2}=35^0\)

\(\widehat{A}=180^0-3\cdot35^0=75^0\)

b: ED//BC

=>\(\widehat{AED}=\widehat{ACB}\)(hai góc đồng vị); \(\widehat{DEB}=\widehat{EBC}\)(hai góc so le trong)

mà \(\widehat{ACB}=\widehat{EBC}\left(=\dfrac{\widehat{ABC}}{2}\right)\)

nên \(\widehat{AED}=\widehat{DEB}\)

=>ED là phân giác của góc AEB