Proving that two line intersect on the circumcircle of the triangle (Solution Shortlist Geometry problem in IMO 2014)
The points P and Q are chosen on the side BC of an acute-angled triangle ABC so that and . The points M and N are taken on the rays AP and AQ, respectively, so that AP = PM and AQ = QN. Prove that the lines BM and CN intersect on the circumcircle of the triangle ABC.
Denote by S the intersection point of the lines BM and CN. Let moreover and . From these equalities it follows that the triangles ABP and CAQ are similar (see Figure 1). Therefore we obtain
Hence the triangles BPM and NQC are similar. This gives , so the triangles BPM and BSC are also similar. Thus we get
which completes the solution.
As in the previous solution, denote by S the intersection point of the lines BM and NC. Let moreover the circumcircle of the triangle ABC intersect the lines AP and AQ again at K and L, respectively (see Figure 2).
Note that and similarly . It implies that the lines BL and CK meet at a point X, being symmetric to the point A with respect to the line BC. Since AP = PM and AQ = QN, it follows that X lies on the line MN.
Therefore, using Pascal’s theorem for the hexagon ALBSCK, we infer that S lies on the circumcircle of the triangle ABC, which ﬁnishes the proof.
Both solutions can be modiﬁed to obtain a more general result, with the equalities
AP = PM and AQ = QN
Also Read : Geometry shortlist problem in IMO 2014