Through the point of intersection of the medians of triangle $ABC$, a line passes, intersecting sides $AB$ and $AC$.

The distances from vertices $B$ and $C$ to this line are $a$ and $b$ respectively. Find the distance from vertex $A$ to this line.
Consider the projections of points $B, A, C$ and the midpoint of side $B C$ onto a given line.

## Solution

Let $M$ be the point of intersection of the medians of triangle $A B C$; $K$ be the midpoint of side $B C$; $E, P, Q$ and $F$ be the projections of points $B, A, K$ and $C$ respectively onto the given line. Since $A K$ is the median of triangle $A B C$, and $M$ is the point of intersection of the medians of this triangle, then $A M: M K=2: 1$.

$K Q$ is the midline of the right trapezoid $B E F C$ (or rectangle, if $b=c$). Therefore,

$$
K Q=\frac{B E+C F}{2}=\frac{b+c}{2}
$$

From the similarity of right triangles $K Q M$ and $A P M$, it follows that

$$
A P=K Q \cdot \frac{A M}{K M}=\frac{b+c}{2} \cdot 2=b+c
$$

## Answer

$b+c$.