Bài 15.Cho tam giác ABC có AB < AC, AD là tia phân giác của $\widehat{BAD}(D\in BC)$....

{
"content1": "Ta có $\widehat{ADB}=\frac{1}{2}(\widehat{CAD}+\widehat{CAD})=\frac{1}{2}(180^{\circ}-\widehat{ACB}+180^{\circ}-\widehat{ABC})=360^{\circ}-\widehat{ACB}-\widehat{ABC}$",
"content2": "Ta cũng có $\widehat{ADC}=\frac{1}{2}\widehat{BAC}=\frac{1}{2}(180^{\circ}-\widehat{ACB}-\widehat{ABC})=180^{\circ}-\widehat{ACB}-\widehat{ABC}$",
"content3": "Vậy $\widehat{ADB}=\widehat{ADC}$ hoặc $\widehat{ADB}<\widehat{ADC}$",
"content4": "Vì $AB}