Do A;B;C;D là 4 góc trong 1 tứ giác \(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
\(\Rightarrow\dfrac{\widehat{A}}{2}+\dfrac{\widehat{B}}{2}+\dfrac{\widehat{C}}{2}+\dfrac{\widehat{D}}{2}=180^0\) (1)
Trong tam giác AEB: \(\widehat{EAB}+\widehat{EBA}+\widehat{AEB}=180^0\)
\(\Leftrightarrow\dfrac{\widehat{A}}{2}+\dfrac{\widehat{B}}{2}+\widehat{AEB}=180^0\) (2)
Trừ vế (1) và (2): \(\Rightarrow\dfrac{\widehat{C}}{2}+\dfrac{\widehat{D}}{2}-\widehat{AEB}=0\Leftrightarrow\widehat{AEB}=\dfrac{\widehat{C}+\widehat{D}}{2}\)
b.
Trong tam giác AFB:
\(\widehat{FAB}+\widehat{FBA}+\widehat{AFB}=180^0\)
\(\Leftrightarrow\dfrac{180^0-\widehat{A}}{2}+\dfrac{180^0-\widehat{B}}{2}+\widehat{AFB}=180^0\)
\(\Leftrightarrow\widehat{AFB}=\dfrac{\widehat{A}+\widehat{B}}{2}\)
