Ta có: $\left \{ {{OA⊥OC} \atop {OB⊥OD}} \right.$ `=>` $\left \{ {{hat(BOD)=90^0} \atop {hat(AOC)=90^0}} \right.$
Đặt: `hat(COD)` là `a` ta có:
$\left \{ {{\hat(BOC)+ \hat(A)=90^0} \atop {\hat(AOD)+\hat(A)=90^0}} \right.$
`=>hat(BOC)=hat(AOD)`
`=>hat(BOC)=hat(AOB)-hat(AOC)=130^0-90^0=40^0`
`=>hat(COD)=130^0-2hat(BOC)=130^0-2*40=50^0`