Butterfly with Menelaus

Sidney Kung
September 7, 2012

We give a proof of the following by using Menelaus' theorem:

Through a point \(P\) of the line segment \(AB\) whose end points \(A\) and \(B\) lie on two intersecting lines \(l\) and \(l'\), respectively, draw \(CD\) and \(EF\) \((C, F\in l\), and \(E,D\in l')\). Let \(CE\cap AB=X\), \(FD\cap AB=Y.\space\)Then \(PA = PB\) implies \(PX = PY.\)

statement of the butterfly theorem in a quadrilateral

Proof

For a repeated application of Menelaus' here, we shall proceed by focusing on parts of the diagram:

  1. \(\triangle OFE\) cut by transversal \(CPD\)

    first step of the proof

    \(\frac{OC}{CF}\times \frac{FP}{PE}\times \frac{ED}{DO} = 1\).

  2. \(\triangle OAB\) cut by transversal \(CPD\)

    second step of the proof

    \(\frac{OD}{DB}\times \frac{BP}{PA}\times \frac{AC}{CO} = 1\).

  3. \(\triangle PBE\) cut by transversal \(FYD\)

    third step of the proof

    \(\frac{PY}{YB}\times \frac{BD}{DE}\times \frac{EF}{FP} = 1\).

  4. \(\triangle FAP\) cut by transversal \(CXE\)

    fourth step of the proof

    \(\frac{PE}{EF}\times \frac{FC}{CA}\times \frac{AX}{XP} = 1\).

Multiplying the four identities and simplifying gives \(\frac{PY}{YB}\times \frac{AX}{XP}\times \frac{BP}{PA} = 1\). Since \(PB=PA\), the latter expression further simplifies to \(\frac{AX}{XP}=\frac{YB}{PY}\). Add now \(1\) to both sides: \(\frac{AX+XP}{XP}=\frac{PY+YB}{PY}\), or \(\frac{PA}{XP}=\frac{PB}{PY},\space\)implying \(PX=PY\).

The above proof moved Hubert Shutrick to make the following observation:

The conics of the pencil through \(C\), \(D\), \(E\), \(F\) intersect the line \(AB\) in pairs of points in involution. Consider the degenerate members \(CD.EF\), \(CE.DF\), and \(l.l'\). \(P\) is a double point because of the first and so, if one of the pairs \(A,B\) and \(X,Y\) is symmetrical about \(P\), then so is the other.

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