cho tam giác ABC có góc B > góc C và góc ngoài BAx. Tia phan giác của góc BAx cắt tia CB tại E.
a. CMR: góc E = \(\dfrac{1}{2}\widehat{B\text{Ax}}-\widehat{C}\)
b. CMR: \(\widehat{E}=\widehat{ABC}-\dfrac{1}{2}\widehat{B\text{Ax}}\)
c. CMR: \(2\widehat{E}=\widehat{ABC}-\widehat{ACB}\)
a: \(\widehat{EAB}=\dfrac{180^0-\widehat{BAC}}{2}=\dfrac{\widehat{ABC}+\widehat{ACB}}{2}\)
\(\widehat{EBA}=180^0-\widehat{ABC}\)
=>\(\widehat{EAB}+\widehat{EBA}=\dfrac{1}{2}\widehat{ABC}+\dfrac{1}{2}\widehat{ACB}+180^0-\widehat{ABC}=-\dfrac{1}{2}\widehat{ABC}+\dfrac{1}{2}\widehat{ACB}+180^0\)
=>\(\widehat{E}=180^0+\dfrac{1}{2}\widehat{ABC}-\dfrac{1}{2}\widehat{ACB}-180^0=\dfrac{1}{2}\widehat{ABC}-\dfrac{1}{2}\widehat{ACB}\)
=>góc E=1/2góc BAx-góc C
b: góc E=1/2góc BAx-góc BAx+góc B
=góc B-1/2góc xAB
c: góc E=1/2góc ABC-1/2góc ACB
=>2*góc E=góc ABC-góc ACB