This answer shows that if we are not allowed to draw extra lines, then there are only four pairs of congruent triangles.
(If we are allowed to draw extra lines, the number of such pairs is more than four since we have $\triangle{OAF}\equiv\triangle{OBG}$.)
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Since we are not allowed to draw extra lines, we have only eight triangles to consider :
$$\triangle{ACO},\quad \triangle{AGO},\quad\triangle{DEG},\quad\triangle{BDO}$$
$$\triangle{BFO},\quad\triangle{DFO},\quad\triangle{CEF},\quad\triangle{CGO}$$
Considering only angles, there are only eight candidates for congruent pairs :
$$(\triangle{ACO},\triangle{GDE}),\quad (\triangle{ACO},\triangle{BDO}),\quad (\triangle{ACO},\triangle{FCE}),\quad (\triangle{AGO},\triangle{BFO})$$
$$(\triangle{GDE},\triangle{BDO}),\quad (\triangle{GDE},\triangle{FCE}),\quad (\triangle{BDO},\triangle{FCE}),\quad (\triangle{DFO},\triangle{CGO})$$
Now, we have
$$\triangle{ACO}\equiv \triangle{BDO},\quad \triangle{GDE}\equiv\triangle{FCE},\quad \triangle{DFO}\equiv\triangle{CGO},\quad\triangle{AGO}\equiv\triangle{BFO}$$
However, we have
$$\triangle{ACO}\not\equiv\triangle{GDE},\quad \triangle{ACO}\not\equiv\triangle{FCE},\quad \triangle{GDE}\not\equiv\triangle{BDO},\quad \triangle{BDO}\not\equiv\triangle{FCE}$$
since
$$AO\not=GE,\quad AC\not=FC,\quad DG\not=DB,\quad BO\not=FE$$
respectively.