Let's define $ABCD$ a trapezoid, $AB \ \| \ CD$ and $AD \perp AB$. Let's consider $M$ a point on the altitude $(AD)$. Given $AB = 6$, $CD = 3$, $AD = 2$, where should be placed $M$ to have $CM + MB$ minimal? (Bulgarian Regional Mathematical Olympiad 1987, 7th grade)
Geometrical approach: Let's draw the circle $\mathcal{C}(C, r = MC)$ with the center $M$, through $C$ and $\mathcal{C} \cap BM = \{B, N\}$. Therefore $BN$ should be minimal to satisfy the condition. However, no clue here.
Algebraic approach: Let's observe that, denoting $AM = x$, we have the expression $$CM + MB = \sqrt{9 + x^2} + \sqrt{36 + (2 - x)^2}$$
Reflect point $C$ over line $AD$ to end up in a new position $C'$ on the line $CD$ extended. Draw $C'M$ and $C'B$.
Now, observe that, $AD$ is the perpendicular bisector of $CC'$ and hence $CM=C'M$. Thus, $CM+BM$ is, by the triangle inequality, least when $M$ is the intersection point of $C'B$ and $AD$. Therefore, the minimum value of $CM+BM$ is $C'B$. Calculate this length by pythagorean theorem now.