I have tried to connect $GO$, $HO$, $IO$, $JO$, $EO$, $AO$, and $FO$, and I know that since the chords are equal, and so as its angles at the center. However, I do not know how to proceed to find the unknown $\angle BAD$ because those points $B$ and $D$ are outside of the circle. Any help will be appreciated.
With your comment, you are very close to the answer. From the power of a point theorem, you have $BG=BF$. Then using $OG=OF$ and $BO$ common side, you can derive that $BO$ is the bisector of $\angle ABC$. Similarly, $DO$ is the bisector of $\angle ADB$. Let's call these angles $2\alpha$ and $2\beta$. Then write the sum of angles in in quadrilaterals $BCDO$ and $BODA$ (they must both be equal to $360^\circ$). $$\alpha+\beta+80^\circ+(360^\circ-160^\circ)=\alpha+\beta+160^\circ+\angle BAD\\\angle BAD=120^\circ$$