Problem:
In convex cyclic quadrilateral ABCD, we know that lines AC and BD intersect at E, lines AB and CD intersect at F, and lines BC and DA intersect at G. Suppose that the circumcircle of △ABE intersects line CB at B and P, and that the circumcircle of △ADE intersects line CD at D and Q, where C,B,P,G and C,Q,D,F are collinear in this order. Prove that if lines FP and GQ intersect at M, then ∠MAC=90∘.
Solution:
First solution. In this particular configuration, we have
hence line AC is the internal angle bisector of angles BAQ and PAD. If we could prove that ∠GAM=∠MAP, then line AM would prove to be the external angle bisector of ∠BAQ and hence perpendicular to AC.
Since △PAF and △QAG are related by ∠PAF=∠QAG, it now suffices to prove that
sin∠MAQsin∠GAM=sin∠MAFsin∠PAM(1)
which is but a repeated application of the Law of Sines. Using the Ratio Lemma in △PAF and △QAG, (1) is equivalent to
MQGM/AQAG=MFPM/AFAP, i.e. MPGM⋅MQFM=AP⋅AQAF⋅AG(2)
We now calculate
However, from △CAP∼△CBE and △CAQ∼△CDE, we have APCP=BECE and AQCQ=DECE. Hence
which gives us (2) and therefore (1). This completes the proof.
Second solution. Note by Power of a Point that CE⋅CA=CP⋅CB=CQ⋅CD. Thus we can perform an inversion at C swapping these pairs of points. The point G is mapped to a point G∗ on ray CB for which QEG∗C is cyclic, but then (using directed angles modulo 180∘ ) we have
∠CG∗E=∠CQE=∠CQP=∠DBC=∠EBC
and so we conclude EB=EG∗. Similarly, ED=EF∗.
Now, M∗, the image of M, is the intersection (distinct from C ) of the circumcircles of △CG∗D and △CF∗B; and we wish to show that ∠EM∗C=90∘.
Note that triangles M∗BG∗ and M∗F∗D are similar, because (again with directed angles)
∠M∗BG∗=∠M∗BC=∠M∗F∗C=∠M∗F∗D
and
∠M∗G∗B=∠M∗G∗C=∠M∗DC=∠M∗DF∗
Then, the same spiral similarity that sends △M∗BG∗ to △M∗F∗D also maps the midpoint K of BG∗ to the midpoint L of F∗D. Consequently, ∠KM∗L=∠BM∗F∗=∠BCF∗=∠KCL, which means that M∗ lies on the circumcircle of triangle KLC as well. In other words, ELCKM∗ is a cyclic pentagon with circumdiameter CE, implying that ∠EM∗C=90∘, as desired.
Third solution. Similarly to the first solution, we begin by noting that
∠GAC=180∘−∠DAC=180∘−∠DBC=∠PBE=180∘−∠PAE.
Thus, AC is the external bisector of ∠GAP. By symmetry, AC is also the external bisector of ∠FAQ.
Now, for a small ϵ>0, consider a homothety of factor 1−ϵ centered at C taking A,G, and Q to A′,G′, and Q′, respectively. Let
X=AP∩A′G′,Y=AF∩A′Q′,M′=PF∩G′Q′
Note that A′G′Q′ and APF are perspective from the point C. Thus, by Desargues' theorem, we know that X,Y, and M′ are collinear.
Moreover, since AC externally bisects ∠GAP and G′A′∣∣GA, it follows that △AXA′ is isosceles, and X lies on the perpendicular bisector of AA′. Similarly, Y also lies on this perpendicular bisector, so the line through M′,X, and Y is perpendicular to AC.
Now, taking ϵ→0, we see that M′→M while X→A and Y→A. It follows that MA is perpendicular to AC, as desired.
