Question

State the dual of the Theorem below.

Let a non-degenerate plane conic touch the sides BC, CA, and AB of a triangle ABC in R² at the points P, Q, and R respectively. Then AP, BQ, and CR are concurrent.

Answer 1

Answer:

Let T be the intersection of the circumcircles of ARQ and BP R.

By collinearity of points, ∠T QA = π − ∠CQT, ∠T RB = π − ∠ART, ∠T P C = π − ∠BP T.

In a cyclic quadrilateral, opposite angles are supplementary.

Therefore ∠T QA = π − ∠ART, ∠T RB = π − ∠BP T.

We conclude ∠T P C = π − ∠CQT.

Now conversely, a quadrilateral whose opposite angles are supplementary is cyclic.

Therefore T also lies on the circumcircle of CQP, as desired.

A defect of the technique is that the relevant theorems depend on the configuration of the points involved, particularly on whether certain points fall between certain other points. For example, one might ask whether the above theorem still holds if P, Q, R are allowed to lie on the extensions of the sides of ABC. It does hold, but the above proof breaks down because some of the angles claimed to be equal become supplementary, and vice versa. Pseudotheorem.:All triangles are isosceles. Pseudoproof. Let ABC be a triangle, and let O be the intersection of the internal angle bisector of A with the perpendicular bisector of BC.

Let D, Q, R be the feet of perpendiculars from O to BC, CA, AB, respectively.

By symmetry across OD, OB = OC, while by symmetry across AO, AQ = AR and OQ = OR.

Now the right triangles ORB and OQC have equal legs OR = OQ and equal hypotenuses OB = OC, so they are congruent, giving RB = QC.

Finally, we conclude AP, BQ, and CR are concurrent.

(Thank you)