A semicircle is constructed outwards on side BC of a triangle ABC as on the diameter. Given points K and L that divide the semicircle into 3 equal arcs, prove that lines AK and AL divide BC into 3 equal parts.

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Solution

Denote the midpoint of side BC by O and the intersection points of AK and AL
with side BC by P and Q, respectively.

We may assume that BP < BQ.

Triangle LCO
is an equilateral one and LC || AB.

Therefore,
△ABQ ∼ △LCQ,

i.e.,
BQ : QC = AB :LC = 2 : 1.

Hence, BC = BQ + QC
= 3QC. Similarly,

BC = 3BP.

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