A line drawn through vertex $C$ of triangle $ABC$ parallel to its bisector $BD$ intersects the extension of side $AB$ at point $M$.

Find the angles of triangle $MBC$, if $\angle ABC=110^{\circ}$.
Use the theorem about angles formed by parallel lines and a transversal.

## Solution

Since $M C \| B D$, then by the property of parallel lines $\angle B M C=\angle A B D=55^{\circ}, \angle B C M=\angle D B C=55^{\circ}$.

## Answer

$55^{\circ}, 70^{\circ}, 55^{\circ}$.