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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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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.

Similarly,

  ∠MQP = ½∠MFP.

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

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