Two Circles and One More

The applet below illustrates the following problem:

  Two circles C(E) with center E and C(F) with center F intersect in points P and Q. EP extended intersects again C(E) in G and C(F) in M. FP extended intersects again C(E) in L and C(F) in H. Then points E, F, L, M, Q are concyclic and ∠LQP = ∠MQP.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

What if applet does not run?

Triangles LEP and MFP are isosceles,right,obtuse,scalene,isosceles with equal base angles EPL and FPM. Therefore, their apex angles are also equal:

  ∠LEP = ∠MFP.

But ∠LEP is the same as ∠LEM and, similarly, ∠MFP is the same as ∠LFM. This makes quadrilateral LEFM cyclic,cyclic,parallelogram,rhombus,trapezoid. We'll show that Q also belongs to the circumcircle LEFM.

∠LQP is inscribed into C(E) and is subtended by the arc of the central,right,acute,central,full angle LEP. Therefore,

  ∠LQP = ½∠LEP.


  ∠MQP = ½∠MFP.

Since ∠LEP = ∠MFP, ∠LQM = ∠LQP + ∠PQM = ∠LEM = ∠LFM. It follows that quadrilaterals LEQM and LQFM are cyclic and their circumcircles coincide.

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

Copyright © 1996-2018 Alexander Bogomolny