Similar Triangles on Sides and Diagonals of a Quadrilateral: What is it about?
A Mathematical Droodle

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Discussion

The applet may suggest the following statement [de Villiers]:

Given any quadrilateral ABCD, if similar triangles ACP, DCQ, and DBR are constructed on AC, DC, and DB, away from AB so that APC = ASB, where S is the intersection of AD and BC extended, then P, Q, R, S are collinear.

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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

ECB = 180° - A - B = APC.

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

PCA + PCD = QCD + PCD.

which implies that DCA = QCP, and therefore,

(1)	CAE = QCP.

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

ACE + ECB = APC + CPQ,

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

180° - A - B = 180° - PAC - DBR,

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

PQD = 180° - CPQ - QCP -DQC = S1CD.

But

BCD = DCA + ACE + ECB

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