Prove that every point, satisfying given condition, lies on some line. Segment $BC$ is the shortest side of noniscosles triangle $ABC$. Points $K$ and $L$ lie respectively on sides $AB$, $AC$ and satisfy condition: for given point $X$ on side $BC$, $BK=BX$ and $CL=CX$. Prove that midpoint of the segment $KL$ lies on some line indepedently of choosing point $X$.
I have been wondering for some time on solution for this problem but I did not come up with anything useful. Could you give some hints?
 A: Note: The condition that $BC$ is the shortest side, or that $ABC$ is non-isosceles, are not necessarily. They might have been introduced to avoid signed lengths and degenerate cases (parallel lines, gradient = 0, etc).
Set this up on the coordinate plane.
Let $ B = (0,0)$, $C = (1,0)$.
Let $D = (1+x_d, y_d)$ be a point on  $AC$ such that $CD = 1$, and $F = ((1+x_d) / 2  , y_d/2) $ be the midpoint of $ BD$.
Let $E = (x_e,y_e)$ be a point on $AB$ such that $EB = 1$, and $G = ( (1+x_e)/2, y_e/2 )$ be the midpoint of $EC$.

Assuming the problem is correct that the locus is a line, we know that the points $F,G$ lies on this line, so it must be the line $ FG$. Let's show this.

Let $X = (t,0)$, then show that

*

*$K = (tx_e, ty_e)$,

*$L  = ((1-t)(x_d) + 1, (1-t)y_d)$.

*The midpoint of $ KL$ is thus $ ( \frac{ tx_e + (1-t)(x_d) + 1 } { 2} , \frac { ty_e  + (1-t)y_d } {2}) $,

*Line $FG$ can be represented by $ ( \frac{1+x_d}{2} , \frac{y_d}{2 }) + t (  \frac{ x_e - x_d}{2}, \frac{y_e - y_d } {2} )$ for some dummy variable $t$.

*The midpoint of $KL$ lies on line $FG$.

Hence, we are done.

Yes, using complex numbers or vectors makes the notation much cleaner, though it's the same.
A: Here's a synthetic solution, which has roots in the coordinate solution.
Let $ D$ be a point on $AC$ such that $ DC  = BC$.
Let $E$ be a point on $AB$ such that $ BE = EC$.
For a given point $X$ on $BC$ such that $ BX : XC = t : 1 - t$, define $Y$ on $ED$ such that $DY : YC = t : 1 - t$.
Show that

*

*$KY \parallel BD \parallel XL$

*$YL \parallel CE \parallel KX$

*$KXLY$ is a parallelogram

*Mid point of $KL$ is the midpoint of $XY$.

*Locus of midpoint of $XY$ is thus a straight line.

*In fact, Midpoint of $XY$ lies on the line connecting the midpoint of $CE$ and $BD$.

Note: To be fair, I'm not certain of a purely synthetic solution to the last statement. It's clearly true by coordinates. I feel that it should be true by some similarity / homothety argument, but that's eluding me for the time being.
A: Let $S,T$ be points of tangency to circle inscribed in triangle $ABC$ respectively on sides $AB$ and $AC$. It is clear that $AK+AL=SA+TA$, meaning that $SK=TL$. Let $D$ be a point on $ST$ such that triangle $DLT$ is iscocles, now $KDLS$ is a parallelogram, therefore midpoint of $KL$ lies on $ST$. This proof described the case when $L$ is on the same side of line $ST$ as point $A$, other is analogous.
