Question

Cho tứ giác ABCD , phân giác góc C và D cắt nhau tại O . Chứng minh rằng :

\(\widehat{COD}=\frac{\widehat{A}+\widehat{B}}{2}\)

Answer 1

ta có A+B=360-(D+C)

<=> A+B=360-2(180-ODC-OCD)=360-360+2.COD=2COD

\(\Rightarrow\widehat{COD}=\frac{\widehat{A}+\widehat{B}}{2}\)

Answer 2

Xét \(\Delta COD\)có :

\(\widehat{COD}=180^o-\left(\widehat{C_1}+\widehat{D_1}\right)\)

\(=180^o-\frac{\widehat{C}+\widehat{D}}{2}\)

xÉT tứ giác ABCD có :

\(\widehat{C}+\widehat{D}=360^o-\left(\widehat{A}+\widehat{B}\right)\)

Do đó : \(\widehat{COD}=180^o-\frac{360^o-\left(\widehat{A}+\widehat{B}\right)}{2}\)

\(\Rightarrow\widehat{COD}=\frac{\widehat{A}+\widehat{B}}{2}\)(đpcm)

Answer 3