# Parallel Lines in a QuadrilateralÁèễ
What is this about?

A Mathematical Droodle

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

Copyright © 1996-2018 Alexander Bogomolny

The applet is intended to suggest the following proposition [Gardiner, Bradley, p. 67]:

ABCD is a quadrilateral. The line through A parallel to CD meets BD at E and the line through D parallel to AB meets AC at F. Prove that EF is parallel to BC.

### Solution

If AB||CD there is nothing to prove since, in this case, E coincides with B and F with C. Otherwise, let M be the intersection of AE and DF and G, H the intersections of EF with AB and CD, respectively. First consider the trapezoid AMCD.

ΔDFH is similar to ΔMFE, so that

(1) | DH/FH = EM/EF, or DH/EM = FH/EF. |

Triangles AEF and CHF are also similar, implying

(2) | CH/AE = FH/EF. |

Together, (1) and (2) give

(3) | DH/EM = FH/EF = CH/AE, or DH/CH = EM/AE. |

In trapezoid ABMD, triangles ABE and MDE are similar. Therefore we have the ratio

(4) | EM/AE = DE/BE. |

Together, (3) and (4) imply

DH/CH = DE/BE. |

Two transversals GH and BC induce equal ratios on sides BD and CD of angle BDC. The transversals are parallel.

Michel Cabart came up with a shorter variant of the proof.

Let I be the intersection of BD and AC. Let g be the homothety with center I mapping BA onto DF, h the homothety with center I mapping DC into EA. Let f the homothety product

We have

f(B) | = hg(B) | = h(D) | = E and also | |

f(C) | = gh(C) | = g(A) | = F |

implying f(BC) = EF so that EF is parallel to BC.

Michel has also observed that by a repeated application of one of *Thales theorems* (the one that claims proportions on parallel transversals which we already used several times above) it is a simple matter to present the same proof while omitting all mention of homotheties:

IE / IB | = (IE / ID) × (ID / IB) | |

= (IA / IC) × (IF / IA) | ||

= IF / IC, |

with the same conclusion as before.

Michel's third message read: "My last remark about this problem. The plain reason why this figure looks so familiar is that it is nothing
else than *Pappus configuration*." You may want to ponder this one before reading further. Just brilliant.

Let *ab*, *cd*, and *bc* be the points at infinity defined by the directions of AB, CD, and BC, respectively. Instead of the quadrilateral ABCD, consider the hexagon BA*ab*CD*cd*. E is then the intersection of A*cd* and CD, while F is the intersection of AC and D*ab*. Let K be the intersection of the third (Pappus) pair, viz., BC and (*ab*)(*cd*). By Pappus' theorem, the three points E, F, K are collinear. But K belongs to the line at infinity (*ab*)(*cd*), implying that EF is indeed parallel to BC.

Note there is another set of parallel lines in a quadrilateral.

### References

- A. D. Gardiner, C. J. Bradley,
*Plane Euclidean Geometry: Theory and Problems*, UKMT, 2005

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

Copyright © 1996-2018 Alexander Bogomolny64851223 |