3.Let $ABC$ be an equilateral triangle, $M$ the midpoint of side $[BC]$, and $D \in (AM)$ such that $\mathrm{AM}+\mathrm{MD}=\mathrm{AB}$. Determine the measure of the angle $\varangle D B M$.


Solution:

![](https://cdn.mathpix.com/cropped/2024_06_07_b45b7a772eefec2dc8bdg-2.jpg?height=563&width=594&top_left_y=278&top_left_x=228)

Let $\mathrm{E} \in ($ AM such that $A E=A B \Rightarrow A M+M E=A B=A M+M D \Rightarrow M E=M D$.

$\triangle A B E$ is isosceles $(A B=A E), m(\angle B A E)=30^{\circ} \Rightarrow m(\angle A B E)=\frac{180-30}{2}=75^{\circ} \Rightarrow m(\angle M B E)$

$$
=75^{\circ}-60^{\circ}=15^{\circ} . \triangle B D M \equiv \triangle B E M(\text { case S.S. }) \Rightarrow m(\angle D B M)=15^{\circ}
$$