# Collinearity via Concyclicity

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A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The applet below provides an illustration to a problem from an outstanding collection by T. Andreescu and R. Gelca:

Let ABC be a triangle and let D and E be points on the sides AB and AC, respectively, such that DE is parallel to BC. Let P be any point interior to triangle ADE and let F and G be the intersections of DE with the lines BP and CP, respectively. Let Q be the second intersection point of the circumcircles of triangles PDG and PFE. Prove that the points A, P, and Q lie on a straight line.

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If A, P, and Q are collinear, there are two pairs of secants to circles (secants ADM/APQ, circle DPG, and secants AEN/APQ, circle PFE). By the Intersecting Secants or Power of a Point theorems (and their inverses), to show that A, P, and Q are collinear, it would suffice to demonstrate that

AD × AM = AE × AN.

This is equivalent to proving that points D, M, E, N are concyclic. However, it is easier to prove that B, M, C, N are concyclic which would imply

AB × AM = AC × AN.

and then employ DE||BC, i.e., AD/AB = AE/AC, to conclude that AD × AM = AE × AN.

Points M, D, P, G are concyclic by the construction. Angles DMP and DGP are subtended by the same arc, so are either equal or supplementary. However, since DE||BC,

**Note** that, as the solution shows, the requirement that P lies in the interior of ΔADE is rather spurious. The construction goes through for P anywhere in the plane, except lines BC and DE; the argument remains valid for all such P.

### References

- T. Andreescu, R. Gelca,
*Mathematical Olympiad Challenges*, Birkhäuser, 2004, 5^{th}printing, 1.3.10 (p. 13)

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny