Trisecting a semicircle gives a trisection of a line segment A semicircle is constructed outwards on side $BC$ of an equilateral triangle $ABC$ as an diameter. Given points $K$ and $L$ which divide the semicircle in $3$ equal arcs. Prove that lines $AK$ and $AL$ divide $BC$ in $3$ equal parts.
 A: Let AK, AL cuts BC at M and N respectively.

The pink triangle is similar to the green triangle and they are in the ratio 2 : 1.
Therefore, CN : NO = 2 :1.
A: Let AB=BC=AC=2.  Let M be the midpoint of BC.  Then MBK,MKL, and MLC are equilateral triangles with sides of length 1.
Let AK cut BC at D  and AL
cut BC at E.  Let P be the midpoint of KL.
Triangle ADE and AKL are similar.  So DE/KL = AM/AP.
KL=1.  AM = $\sqrt {3}.$ MP = $\sqrt {3}/2$.  So AP = $\sqrt {3}(1+1/2) $
Solving DE = 2/3. Or 1/3 of BC.
ADE is clearly isoceles.  So DE is central to BC.
So AD=DE=EC=2/3= 1/3 of BC.
A: Extend $BK$ and $CL$ until they intersect at their common point of intersection $A^*$. Since the three arcs are equal, $\angle \, BCA^* = \angle \, BCL = \angle \, CBL = \angle \, CBA^* = \frac{2}{3} \big(\frac{1}{2}\text{arc}(BKLC) \big) = \frac{2}{3} 90^{\circ} = 60^{\circ}$. Therefore triangle $A^*BC$ is equilateral and congruent to $ABC$.  Furthermore, $C\angle \, CLB = \angle \, BKC = 90^{\circ}$ because $BC$ is the diameter of the arc $\text{arc}(BKLC)$, which means that $L$ and $K$ are midpoints of $CA^*$ and $BA^*$ respectively. Let $M$ be the intersection point of $AA^*$ and $Bc$. Then since $ABC$ and $A^*BC$ are congruent equilateral triangles sharing a common edge $BC$, the quadrilateral $ABA^*C$ is a rhombus, so $M$ is the midpoint of both $AA^*$ and $BC$  and $AA^*$ is perpendicular to $BC$. Consequently, $AA^*B$ and $AA^*C$ are congruent isosceles triangles. Since $BM$ and $AK$ are medians in $AA^*B$, their intersection point $G_B$ is the centroid of $AA^*B$. Since $CM$ and $AL$ are medians in $AA^*C$, their intersection point $G_C$ is the centroid of $AA^*C$. By congruence of $AA^*B$ and $AA^*C$, segments $BG_B = CG_C$ and $G_BM = \frac{1}{2}BG_B = \frac{1}{2}CG_C = G_CM$ and thus $G_BG_C = BG_B =CG_C$.   
