Similar Triangles on Sides and Diagonals of a Quadrilateral
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The applet may suggest the following statement [de Villiers]:

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

1. One does not require AD = BC;
2. One does not require ∠BAD + ∠CBA = 120°
3. 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, ∠CAE = ∠DCA and ∠CFB = ∠A. In ΔCFB we therefore have

From the similarity of triangles PAC and QDC, we have ∠PCA = ∠QCD. Therefore

which implies that ∠DCA = ∠QCP, and therefore,

From the similarity of triangles PAC and QDC we also have PC/AC = QC/DC. But DC = AE since ADCE is a parallelogram, and therefore

By (1) and (2), triangles AEC and CQP are similar, which implies ∠ACE = ∠CPQ. Since ∠APC = ∠ECB, we have

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

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 ∠ACB = ∠APQ, we have ∠AC'B' = ∠APQ and, therefore, C'B' is parallel to PQ. Thus PQ is inclined to BC at an angle of size ∠PAC. Similarly, from a clockwise rotation around B through ∠DBR that takes A to A' and D to D', we have A'D' is parallel to QR, that is, QR is inclined to AD at an angle of -∠DBR.

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 ∠QS₁D = ∠QCD with S₁ on QR. Connect S₁ with C. We now prove that ∠DS₁C = 180° - ∠A - ∠B and that S₁CD and S₁DA are straight lines, hence S₁ and S coincide.

From the construction we have QDCS₁ is a cyclic quadrilateral. Thus ∠DS₁C = ∠DQC = 180°. Also, ∠S₁CD = ∠PQD.

In ΔPQC, we have

But

and ∠DCA = ∠QCP, ∠ACE = ∠CPQ and ∠ECB = ∠DQC.

Therefore ∠S₁CD + ∠BCD = 180° and S₁CB is a straight line. Similarly we can prove that S₁DA is a straight line. Therefore, indeed, S₁ and S are one and the same point, namely the intersection of AD and BC. (Note that if S falls on QP, we simply ∠QS₁C = ∠QDC and show in the same manner that S₁ and S coincide.)

References

1. 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

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