Answer:
The correct option is;
CGB
Whereby, we have;
Triangle RKB ~ Triangle CGB
Explanation:
The given triangles have the following congruent angles;
Angle ∠RKB on triangle ΔRKB is congruent to angle ∠CGB on triangle ΔCGB
We are given angle ∠KBR on triangle ΔRKB is congruent to angle ∠GBC on triangle ΔCGB
Therefore, ΔRKB is similar to ΔCGB by the Angle-Angle (AA) rule for similarity.