ta có A+B=360-(D+C)

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

=>COD=(A+B)/2

ta có A+B=360-(D+C)

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

=>COD=(A+B)/2

Em ms lp 7!!!

Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

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

\(\widehat{COD}=180^o-\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)\)

\(=180^o-\frac{1}{2}\left[360^o-\left(\widehat{A}+\widehat{B}\right)\right]\)

\(=\frac{1}{2}\left(\widehat{A}+\widehat{B}\right)\)

ta có:góc DOC=180^o-(góc ODC+góc OCD)

                          =180^o-\(\frac{gócD+gócC}{2}\)

                          =180^o-\(\frac{360^o-\left(gócA+gócB\right)}{2}\)

                           =180^o-180^o+\(\frac{gócA+gócB}{2}\)

                          =\(\frac{gócA+gócB}{2}\)

bn ơi mình giải đk nhưng mà mình ko bít kí hiệu nên làm sẽ hơi khó nhìn, thông cảm nha

ko sao,bn cứ làm đi

1: Xét ΔIAB có \(\hat{AIB}+\hat{IAB}+\hat{IBA}=180^0\)

=>\(\hat{IAB}+\hat{IBA}=180^0-65^0=115^0\)

=>\(\frac12\left(\hat{BAD}+\hat{ABC}\right)=115^0\)

=>\(\hat{BAD}+\hat{ABC}=115^0:\frac12=230^0\)

Xét tứ giác ABCD có \(\hat{BAD}+\hat{ABC}+\hat{BCD}+\hat{CDA}=360^0\)

=>\(\hat{BCD}+\hat{CDA}=360^0-230^0=130^0\)

mà \(\hat{BCD}-\hat{CDA}=10^0\)

nên \(\hat{BCD}=\frac{130^0+10^0}{2}=70^0;\hat{CDA}=70^0-10^0=60^0\)