[alogo] Equal segments division-2

Find points I, K on the sides of a triangle ABC, such that segments CI, IK, KB are equal in length.
Just another solution from that exposed in EqualSegments.html .

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]
[2_0] [2_1] [2_2]

Solution taken from the book of Yaglom.
It relies on the remark that IC, KB being equal, there is a rotation about some point D that brings the one to the other. D can be determined without knowing the exact lengths of CI, IK, etc. In fact, the center of the rotation has to be on the medial line of BC and triangles DCI, DBK must be equal, the angles at D being angle(CDB) = angle(IDK) = angle(CAB). Thus D is one of the intersection points of the circumcircle of ABC with the medial line of BC.
Now the isosceles DCB with a known angle at D has a known ratio of side-lengths (depending on angle(CAB)) k = CI/ID = IK/ID = CB/DC = CB/DB. Hence I lies on the Apollonian circle of segment CD, which is the locus of points P, such that PC/PD = k.

See Also

EqualSegments.html

References

Yaglom, I. M. Geometric Transformations I(3 vols.) Washington DC, Math. Assoc. Ammer., 1962, pp. 133.

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