1. **Problem Statement:** In triangle $\triangle ABC$, $BM$ and $CN$ are perpendiculars from vertices $B$ and $C$ respectively onto any line passing through $A$. $L$ is the midpoint of segment $BC$. Prove that $ML = NL$. 2. **Understanding the problem:** We have a triangle $ABC$ and a line through $A$. From $B$ and $C$, perpendiculars $BM$ and $CN$ are dropped onto this line. $L$ is the midpoint of $BC$. We need to prove that the distances from $M$ and $N$ to $L$ are equal. 3. **Key concepts:** - Midpoint divides a segment into two equal parts. - The perpendiculars from $B$ and $C$ to the same line through $A$ create right angles. - Using coordinate geometry or vector approach can simplify the proof. 4. **Proof using coordinate geometry:** - Place $A$ at the origin $(0,0)$. - Let the line through $A$ be along the $x$-axis for simplicity. - Coordinates of $B$ and $C$ are arbitrary: $B=(x_1,y_1)$, $C=(x_2,y_2)$. - Since $BM$ and $CN$ are perpendiculars to the $x$-axis, their feet $M$ and $N$ have coordinates $M=(x_1,0)$ and $N=(x_2,0)$. - The midpoint $L$ of $BC$ is $L=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. 5. **Calculate distances $ML$ and $NL$:** - Distance $ML = \sqrt{\left(x_1 - \frac{x_1+x_2}{2}\right)^2 + \left(0 - \frac{y_1+y_2}{2}\right)^2}$ - Distance $NL = \sqrt{\left(x_2 - \frac{x_1+x_2}{2}\right)^2 + \left(0 - \frac{y_1+y_2}{2}\right)^2}$ 6. **Simplify inside the square roots:** - $ML = \sqrt{\left(\frac{x_1 - x_2}{2}\right)^2 + \left(\frac{y_1 + y_2}{2}\right)^2}$ - $NL = \sqrt{\left(\frac{x_2 - x_1}{2}\right)^2 + \left(\frac{y_1 + y_2}{2}\right)^2}$ 7. **Note that $(x_1 - x_2)^2 = (x_2 - x_1)^2$, so:** - $ML = NL$ 8. **Conclusion:** The distances from $M$ and $N$ to $L$ are equal, i.e., $ML = NL$. This completes the proof.