Ta có: \(AD=AB\left(gt\right)\Rightarrow\Delta ABD\) cân tại A

\(\Rightarrow\widehat{ABD}=\widehat{ADB}=\dfrac{180^0-\widehat{BAD}}{2}\)

Mà \(\widehat{BAD}=\widehat{ABC}+\widehat{ACB}\) ( tính chất góc ngoài của tam giác)

\(\Rightarrow\widehat{ABD}=\dfrac{180^0-\left(\widehat{ABC}+\widehat{ACB}\right)}{2}\)

Ta có: \(\widehat{CBD}=\widehat{ABD}+\widehat{ABC}=\dfrac{180^0-\left(\widehat{ABC}+\widehat{ACB}\right)}{2}+\widehat{ABC}=\dfrac{180^0-\widehat{ABC}-\widehat{ACB}+2\widehat{ABC}}{2}=\dfrac{180^0+\left(\widehat{ABC}-\widehat{ACB}\right)}{2}=90^0+\dfrac{\alpha}{2}\)

linh nhão l ơi

Bài làm

Xét tam giác ABC có:

\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^0\)

=> \(\widehat{BAC}=180^0-\widehat{ABC}-\widehat{ACB}\)

=> \(\widehat{BAC}=180^0-\widehat{ABC}-\widehat{ACB}\)

Mà \(\widehat{BAC}=\widehat{ABC}-\widehat{ACB}\)

=> \(180^0-\widehat{ABC}-\widehat{ACB}=\widehat{ABC}-\widehat{ACB}\)

=> \(180^0-\widehat{ABC}-\widehat{ABC}-\widehat{ACB}+\widehat{ACB}=0\)

=> \(180^0-2\widehat{ABC}=0\)

=> \(2\widehat{ABC}=180^0\)

=> \(\widehat{ABC}=180^0:2=90^0\)

Ai làm được nốt không? 

??????????????

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\widehat{BAD}=\widehat{B}+\widehat{C}\)

\(\widehat{ABD}=\frac{180^o-\widehat{BAD}}{2}=90^o-\frac{\widehat{B}+\widehat{C}}{2}\)

\(\widehat{CBD}=\widehat{B}+\widehat{ABD}=\widehat{B}+90^o-\frac{\widehat{B}+\widehat{C}}{2}=90^o+\frac{\widehat{B}-\widehat{C}}{2}=90^o+\frac{\alpha}{2}\)

tui ko bít làm

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