`\hat{A}+\hat{B}+\hat{C}=180^0` (tổng ba góc trong tam giác

`\hat{AKD}+\hat{DKI}=180^0` (góc bẹt)

`\hat{AKE}+\hat{EKI}=180^0` (góc bẹt)

`⇒\hat{AKD}+\hat{DKI}+\hat{AKE}+\hat{EKI}=180^0+180^0=360^0`

`⇒\hat{EKD}+\hat{DKI}+\hat{EKI}=360^0`

`⇒\hat{EKD}+\hat{DKI}+\hat{EKI}+\hat{A}+\hat{B}+\hat{C}=360^0+180^0=540^0`

Đáp án:

 BẠN THAM KHẢO NHA!

Giải thích các bước giải:

 Vì $\widehat{A}; \widehat{B};\widehat{C}$ là 3 góc trong của $ΔABC$

$⇒\widehat{A}+\widehat{B}+\widehat{C}=180^o$

Lại có: $\widehat{EKD}; \widehat{EKI};\widehat{IKD}$ là 3 góc kề nhau 

$⇒\widehat{EKD}+\widehat{EKI}+\widehat{IKD}=360^o$

$⇒\widehat{A}+\widehat{B}+\widehat{C}+\widehat{EKD}+\widehat{EKI}+\widehat{IKD}=180^o+360^o$

$⇒\widehat{A}+\widehat{B}+\widehat{C}+\widehat{EKD}+\widehat{EKI}+\widehat{IKD}=540^o$