# Proving that two line intersect on the circumcircle of the triangle (Solution Shortlist Geometry problem in IMO 2014)

Problem 1

The points P and Q are chosen on the side BC of an acute-angled triangle ABC so that $\angle PAB=\angle ACB$ and $\angle QAC=\angle CBA$. 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.

Solution 1.

Denote by S the intersection point of the lines BM and CN. Let moreover $\beta =\angle QAC =\angle CBA$ and $\gamma =\angle PAB =\angle ACB$. From these equalities it follows that the triangles ABP and CAQ are similar (see Figure 1). Therefore we obtain $\frac{BP}{PM}=\frac{BP}{PA}=\frac{AQ}{QC}=\frac{NQ}{QC}$

Moreover, $\angle BPM=\beta +\gamma =\angle CQN$.

Hence the triangles BPM and NQC are similar. This gives $\angle BMP=\angle NCQ$, so the triangles BPM and BSC are also similar. Thus we get $\angle CSB=\angle BPM=\beta +\gamma =180^0-\angle BAC$,

which completes the solution. Solution 2.

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 $\angle LBC=\angle LAC=\angle CBA$ and similarly $\angle KCB=\angle KAB=\angle BCA$. 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.

Comment.

Both solutions can be modiﬁed to obtain a more general result, with the equalities

AP = PM and AQ = QN

replaced by $\frac{AP}{PM}=\frac{QN}{AQ}$.

Also Read : Geometry shortlist problem in IMO 2014