## Similar Triangles on Sides and Diagonals of a Quadrilateral

What is it about?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2018 Alexander BogomolnyThe applet may suggest the following statement [de Villiers]:

What if applet does not run? |

This is a generalization of a statement concerning equilic quadrilaterals; the relaxation being three-fold:

- One does not require AD = BC;
- One does not require ∠BAD + ∠CBA = 120°
- The constructed triangles are similar and not necessarily equilateral.

Connect P with Q and Q with R. Construct CE parallel and equal to AD as in the applet. Let CE cuts AB in point F. Since ADCE is a parallelogram,

From the similarity of triangles PAC and QDC, we have

which implies that

(1) | ∠CAE = ∠QCP. |

From the similarity of triangles PAC and QDC we also have

(2) | PC/AC = QC/AE. |

By (1) and (2), triangles AEC and CQP are similar, which implies

and so ∠ACB = ∠APQ. By constructing DG parallel and equal to CB, we can prove in a similar manner that

A counterclockwise rotation of size ∠PAC around A carries C to a point C' on the line through AP, and B to B'. But, since

Since AD and BC are inclined towards each other at an angle of

rotating BC through ∠PAC and AD through -∠DBR (in appropriate direction), aligns B'C' and A'D' in the same direction, and we therefore have that PQ and QR also line up in the same direction, i.e., P, Q, and R are collinear.

Next construct ∠QS_{1}D = ∠QCD with S_{1} on QR. Connect S_{1} with C. We now prove that _{1}C = 180° - ∠A - ∠B_{1}CD and S_{1}DA are straight lines, hence S_{1} and S coincide.

From the construction we have QDCS_{1} is a cyclic quadrilateral. Thus _{1}C = ∠DQC = 180°._{1}CD = ∠PQD.

In ΔPQC, we have

_{1}CD.

But

and ∠DCA = ∠QCP, ∠ACE = ∠CPQ and

Therefore ∠S_{1}CD + ∠BCD = 180° and S_{1}CB is a straight line. Similarly we can prove that S_{1}DA is a straight line. Therefore, indeed, S_{1} and S are one and the same point, namely the intersection of AD and BC. (Note that if S falls on QP, we simply _{1}C = ∠QDC_{1} and S coincide.)

### References

- M. de Villiers,
__The Role of Proof in Investigative, Computer-based Geometry: Some Personal Reflections__, in*Geometry Turned On*, MAA Notes 41, 1997, pp. 15-24

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2018 Alexander Bogomolny70776381