1. **Stating the problem:** We have an obtuse angle $x\hat{O}y$ with vertex $O$. The bisector of this angle is the semi-line $[Ou]$. Points $F$ and $R$ lie on $[OY]$ and $[OX]$ respectively, with $OF=6$ cm and $OR=3$ cm. $E$ is the midpoint of $[OF]$. Point $M$ lies on $[Ou]$. We need to show that triangles $RMO$ and $OME$ are congruent and then deduce the type of triangle $RME$. 2. **Drawing the figure:** - $O$ is the origin. - $R$ is on $OX$ at 3 cm from $O$. - $F$ is on $OY$ at 6 cm from $O$. - $E$ is midpoint of $OF$, so $OE=\frac{OF}{2}=3$ cm. - $M$ lies on the bisector $[Ou]$ of the angle $x\hat{O}y$. 3. **Showing triangles $RMO$ and $OME$ are congruent:** - Since $[Ou]$ is the bisector of angle $x\hat{O}y$, it divides the angle into two equal angles. - Therefore, $\angle ROM = \angle EOM$ because $M$ lies on the bisector. - $OM$ is common side to both triangles $RMO$ and $OME$. - $OR = OE = 3$ cm (since $E$ is midpoint of $OF$ and $OR=3$ cm). 4. **Using the Side-Angle-Side (SAS) congruence criterion:** - Triangles $RMO$ and $OME$ have two sides and the included angle equal: $$OR = OE$$ $$\angle ROM = \angle EOM$$ $$OM = OM$$ - Hence, $\triangle RMO \cong \triangle OME$. 5. **Deducing the type of triangle $RME$:** - Since $\triangle RMO \cong \triangle OME$, corresponding sides are equal: $$RM = ME$$ - Therefore, triangle $RME$ has two equal sides $RM$ and $ME$. - Hence, $\triangle RME$ is isosceles. **Final answer:** - Triangles $RMO$ and $OME$ are congruent by SAS. - Triangle $RME$ is isosceles because $RM = ME$.