Question

[Nguyễn Thị Đoan Trang]

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: \(2\widehat{E}=\widehat{ABC}-\widehat{ACB}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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