Đáp án: `\hat{A}=140^o;\hat{B}=90^o`

 

Giải thích các bước giải:

 Vì `DI` là p/g `\hat{ADC}`

`=>\hat{IDC}=1/2\hat{ADC}`

`=>\hat{ADC}=2\hat{IDC}`

C/m tương tự : `=>\hat{BCD}=2\hat{ICD`

Xét `ΔIDC` có : `\hat{IDC}+\hat{ICD}+115^o=180^o`

`=>\hat{IDC}+\hat{ICD}=65^o`

`=>2\hat{IDC}+2\hat{ICD}=130^o`

Vì `\hat{A}-\hat{B}=50^o`

`=>\hat{A}=50^o+\hat{B`

Xét tứ giác `ABCD` có :

`\hat{A}+\hat{B}+\hat{ADC}+\hat{BCD}=360^o`

`=>50^o+\hat{B}+\hat{B}+2\hat{ICD}+2\hat{IDC}=360^o`

`=>50^o+2\hat{B}+130^o=360^o`

`=>2\hat{B}=180^o`

`=>\hat{B}=90^o`

`=>\hat{A}=90^o +50^o=140^o`

Vậy `\hat{A}=140^o;\hat{B}=90^o`

Xét $ΔCID$:

$CID+IDC+ICD=180^o$

⇒ $IDC+ICD=180^o-115^o=65^o$

⇒ $2.IDC+2.ICD=2.65^o$

⇒ $D+C=130^o$

Xét tứ giác $ABCD$ ⇒ $A+B+C+D=360^o$

⇒ $A+B=360^o-130^o=230^o$

⇒ $A=230^o-B$

⇒ $(230^o-B)-B=50^o$

    $230^o-B-B=50^o$

    $230^o-2B=50^o$

    $2B=230^o-50^o=180^o$

⇒ $B=180^o:2=90^o$

⇒ $A=230^o-90^o=140^o$

Vậy $A=140^o$, $B=90^o$