## Subject 3.

Consider the triangle $ABC$ where $m(\widehat{ABC})=30^{\circ}$ and $m(\widehat{ACB})=15^{\circ}$. Point $M$ is the midpoint of side $[BC]$. Determine the measure of the angle $\widehat{AMB}$.

| Details of solution | Associated grading |
| :--- | :---: |
| Consider the ray (Cx such that ray (CA is the angle bisector of $\widehat{BCx}$. Let $\{F\}=AB \cap (Cx$. Triangle $FBC$ is isosceles with base $[BC]$, so $FM \perp BC$ and $m(\widehat{BFM})=m(\widehat{CMF})=m(\widehat{BFx})=60^{\circ}$. | $\mathbf{3 p}$ |
| For triangle $CMF$, ray (CA is the internal angle bisector, and ray (FA is the external angle bisector. We deduce that ray (MA is also the external angle bisector for triangle $CMF$. | $\mathbf{3 p}$ |
| It follows that $m(\widehat{AMB})=45^{\circ}$. | $\mathbf{1 p}$ |
3. Consider triangle $ABC$ where $m(\widehat{ABC})=30^{\circ}$ and $m(\widehat{ACB})=15^{\circ}$. Point $M$ is the midpoint of side $[BC]$. Determine the measure of angle $\widehat{AMB}$.