### Two quadrilaterals: What Is This About?

A Mathematical Droodle

What if applet does not run? |

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

Copyright © 1996-2018 Alexander Bogomolny

The configuration of a quadrilateral A'B'C'D' circumscribed around a circle with points of tangency A, B, C, and D forming another quadrilateral ABCD has many engaging properties. For example, the diagonals of the two quadrilaterals meet at the same point. We also know that the four points of intersections of the opposite sides of two quadrilaterals are collinear. Buried in the proof of the latter statement is another property which is of interest in its own right:

What if applet does not run? |

### Theorem 1

The point of intersection M of the sides AB and CD lies on the diagonal B'D'; the point of intersection K of sides AD and BC lies on the diagonal A'C'.

### Proof

The proof exploits the well known theorem of Menelaus. Let in ΔA'B'D' the transversal AB crosses B'D' in point M. Then

(1) | B'M/MD' · D'A/AA' · A'B/BB' = -1, |

which, since AA' = A'B, reduces to

(2) | B'M/MD' = - BB'/D'A. |

Similarly, assume the transversal CD in ΔB'C'D' crosses B'D' in some other point S (not shown). Then similarly to (1) we obtain

(3) | B'S/SD' · D'D/DC' · C'C/CB' = -1; |

and, as a consequence,

(4) | B'S/SD' = - CB'/D'D. |

Comparison of (2) and (4) shows that B'M/MD' = B'S/SD', so that M and S are one and the same point. Thus indeed AB and CD meet on B'D' (in M). In the same manner, BC and AD meet on A'C'.

Note that from what we've just seen, if, say, AB is parallel to B'D' then necessarily also CD||B'D', in which case the three meet at a point in infinity and the assertion of Theorem 1 still holds.

In fact more is true. The above statement admits a converse. Assume a convex quadrilateral ABCD is inscribed (one vertex per side) into another convex quadrilateral A'B'C'D'.

### Theorem 2

Then the side lines AB and CD meet on B'D' iff the side lines AD and BC meet on A'C'.

### Proof

As it follows from (1) and (3), AB and CD meet in M on B'D', iff

(5) | AA'·BB' / A'B·D'A = CB'·DC' / C'C·D'D. |

Repeating the same steps for the transversals BC and AD shows that (5) is exactly the condition for BC and AD to meet on A'C', which proofs Theorem 2.

### References

- R. Honsberger,
*Mathematical Chestnuts from Around the World*, MAA, 2001, pp 98-101

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

Copyright © 1996-2018 Alexander Bogomolny

72018444