Answer
The reaction is formation of carbenes by alpha elimination. It forms a cyclopropane.
Work Step by Step
The proposed mechanism is:
Step 1: The $OH^-$ which comes from a strong base attacks the H in $CHBr_3$. The electrons in the bond between C-H go to the C.
Step 2: The pi bonds in the double bond attack on the $CBr_2$ and the lone pair on the $CBr_2$ attack on the double bond, forming a cyclopropane across the double bond.