Question

Answer 1

Xét t/g `ABCD` có: `\hat{BAD}+\hat{ABC}=360^o -100^o -60^o =200^o`

Vì `AI;BI` là tia p/g của `\hat{BAD};\hat{ABC}=>{(\hat{A_2}=1/2\hat{BAD}),(\hat{B_2}=1/2\hat{ABC}):}`

Xét `\triangle AIB` có: `\hat{A_2}+\hat{B_2}+\hat{AIB}=180^o`

    `=>1/2\hat{BAD}+1/2\hat{ABC}+\hat{AIB}=180^o`

    `=>1/2(\hat{BAD}+\hat{ABC})+\hat{AIB}=180^o`

   `=>1/2 .200^o+\hat{AIB}=180^o`

   `=>\hat{AIB}=80^o`