In a triangle ABC, let L, M be the midpoints of the sides BC, CA respectively. Prove that AL = BM if and only if AC = BC I've managed to prove that if AC = BC, then AL = BM by observing that for triangles ABL and ABM, since AB is shared and AM = BL and $\angle$ BAM = $\angle$ ABL, since triangle ABC is isoceles, by SAS congruence axiom, triangles ABL and ABM are congruent, therefore AL = BM. I'm having trouble proving the opposite statement. Any help would be appreciated
 A: Let $G$ be the intersection of $AL$ and $BM$. The point $G$ is the centroid of the triangle, so $AG=2GL$ and $BG=2GM$. Now, using $AL=BM$, proceed as follows:

*

*Prove that $\triangle AGM \cong \triangle BGL$;

*Then, that $\triangle BGC \cong \triangle AGC$.

From the second congruence it follows that $AC=BC$.
A: Let
$|AL|=m_a$,
$|BM|=m_b$,
$|BC|=a$,
$|AC|=b$,
$|AB|=c$.
Using known expression of the length of median in terms of the side lengths of triangle,
\begin{align}
m_a&=\tfrac12\sqrt{2b^2+2c^2-a^2}
,\\
m_b&=\tfrac12\sqrt{2a^2+2c^2-b^2}
,
\end{align}
$m_a=m_b$ implies
\begin{align}
2b^2+2c^2-a^2
&-
(2a^2+2c^2-b^2)
=0
,\\
3(b+a)(b-a)&=0
\end{align}
and $a=b$ follows.
A: $$\overrightarrow{CL}=\frac{1}{2}
\overrightarrow{CB},\quad
\overrightarrow{CM}=\frac{1}{2}
\overrightarrow{CA},$$
$$\overrightarrow{BM}=\overrightarrow{CM}-\overrightarrow{CB}=
\frac{1}{2}\overrightarrow{CA}-\overrightarrow{CB},\quad
\overrightarrow{AL}=\overrightarrow{CL}-\overrightarrow{CA}=
\frac{1}{2}\overrightarrow{CB}-\overrightarrow{CA},$$
$$|\overrightarrow{AL}|=|\overrightarrow{BM}|\Leftrightarrow
|\overrightarrow{AL}|^2=|\overrightarrow{BM}|^2\Leftrightarrow
\overrightarrow{AL}^2=\overrightarrow{BM}^2\Leftrightarrow
$$
$$
\left(
\frac{1}{2}\overrightarrow{CB}-\overrightarrow{CA}
\right)^2=
\left(
\frac{1}{2}\overrightarrow{CA}-\overrightarrow{CB}
\right)^2\Leftrightarrow\tag{1}
$$
$$
\frac{1}{4}\overrightarrow{CB}^2
-\overrightarrow{CB}\cdot\overrightarrow{CA}
+\overrightarrow{CA}^2=
\frac{1}{4}\overrightarrow{CA}^2
-\overrightarrow{CA}\cdot\overrightarrow{CB}
+\overrightarrow{CB}^2\Leftrightarrow\tag{2}
$$
$$
\frac{3}{4}\overrightarrow{CA}^2=
\frac{3}{4}\overrightarrow{CB}^2\Leftrightarrow\tag{3}$$ $$
\overrightarrow{CA}^2=\overrightarrow{CB}^2\Leftrightarrow
|\overrightarrow{CA}|^2=|\overrightarrow{CB}|^2\Leftrightarrow
|\overrightarrow{CA}|=|\overrightarrow{CB}|,\hbox{ QED.}
$$

This problem excellently fits the vectors approach.
Generally, we don't need a picture when solving with vectors, however, step $1$ of "1. Select the most convinient basis. 2. Express everything given and needed via the basis vectors 3. Solve algebraically 4. Don't think" may require a picture.

We see that most convinient here is to take basis vectors $a:=\overrightarrow{CA}$ and $b:=\overrightarrow{CB}$ because of symmetry. And we have
$$\left(\frac12 a-b\right)^2=\left(\frac12 b-a\right)^2\Leftrightarrow\\
\frac14 a^2-ab+b^2=\frac14 b^2-ab+a^2\Leftrightarrow\\
\frac14 a^2+b^2=\frac14 b^2+a^2\Leftrightarrow
\frac34 b^2=\frac34 a^2$$
and we're done.
Basically, it's the same solution as the g.kov's answer except for that we don't have to memorize the median length formula.
