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?

• By setting $X = B, X = C$, you can determine the (supposed) line. What can we say about this line? Jan 24 '21 at 20:52
• Hint: Use complex numbers (assuming you're familiar with it) and the result follows easily. Jan 24 '21 at 21:06
• @CalvinLin It is chord of semicircle $BC$. Only elementary geometry is allowed Jan 24 '21 at 21:07
• sure, use coordinate geometry and bash it through. Jan 24 '21 at 21:08

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.

• Thank you for your answer. But saying "only elementary geometry" I also meant no coordinate geometry Jan 24 '21 at 21:38
• @1qwertyyyy Then you should say "using synthetic / axiomatic / pure geometry" instead. I do consider coordinate geometry as elementary. Jan 24 '21 at 21:39
• Thank you for correction, I'm sorry for misunderstanding Jan 24 '21 at 21:44

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.