In $\triangle A B C, D$ and $E$ are two mid points of sides $A B$ and $\mathrm{AC}$ respectively. If $\angle \mathrm{BAC}=40^{\circ}$ and $\angle \mathrm{ABC}=$ $65^{\circ}$ then $\angle \mathrm{CED}$ is :

$\angle \mathrm{BAC}=40^{\circ}$

$\angle \mathrm{ABC}=65^{\circ}$

$\therefore \angle \mathrm{ACB}=180^{\circ}-40^{\circ}-65^{\circ}=75^{\circ}$

$\mathrm{DE} \| \mathrm{BC}$

$\therefore \quad \angle \mathrm{AED}=\angle \mathrm{ACB}=75^{\circ}$

$\therefore \quad \angle \mathrm{CED}=180^{\circ}-75^{\circ}=105^{\circ}$