Angle chasing might not be the most advanced technique in geometry, but it can be quite fun. And it still takes some spotting.
Here is what I got by playing on GeoGebra again:
A triangle ABC has circumcircle Γ. The internal angle bisector of BCA meets Γ at D. The internal angle bisectors of BDC and CDA meet Γ again at E and F respectively. DE, DC and DF cut AB at J, K and L respectively. Let EC and DF be extended to meet at X. There are 3 things I noticed:
- EFLJ is cyclic
- ECKJ is cyclic
- The circumcentre of DCX lies on DA (possibly extended)
Since it looks like such a mess, we’ll prove them one by one.
Proof of 1.
For simplicity, a, b and c will denote the angles ∠BAC, ∠ABC and ∠ACB respectively.
∠ABC = ∠ADC (same segment) so ∠FDA = b/2
Also ∠BAD = ∠BCD = c/2 (same segment)
By angles in a triangle, ∠DLA = 180 – b/2 – c/2
∠DLA = ∠JLF (vertically opposite)
Now ∠DEA = ∠DCA = c/2 (same segment)
Also ∠AEF = ∠FDA = b/2 (same segment)
So ∠JEF = ∠DEA + ∠AEF = b/2 + c/2
Hence ∠JEF + ∠JLF = 180, so by the converse of ‘opposite angles in a cyclic quadrilateral are suppplementary’, EFLJ is cyclic.
Proof of 2.
∠BAC = ∠BDC = a
As ∠BCD = c/2, ∠DBC = 180 – a – c/2 by angles in a triangle
∠DEC = ∠DBC (same segment)
Also recall that ∠ADC = b and ∠BAD = c/2
So by angles in a triangle, ∠DKA = 180 – b – c/2
∠JKC = ∠DKA (vertically opposite)
∠JEC + ∠JKC = 180 – a – c/2 + 180 – b – c/2 = 360 – (a + b + c)
But a + b + c = 180 (angles in a triangle)
So ∠JEC and ∠JKC are supplementary; so again, ECKJ is cyclic by the converse of ‘opposite angles in a cyclic quadrilateral are suppplementary’.
Of course, CFLK is also cyclic for the same reasons.
Edit: Faster Proof
∠ECD = ∠EFD (same segment) ie ∠ECK = ∠EFL
But ∠EJK + ∠EFL = 180 (opposite angles in cyclic quadrilateral EFLJ)
So ∠EJK + ∠ECK = 180; hence ECKJ is cyclic
Proof of 3.
Let the circumcircle of DCX be Λ. Extend DA to meet Λ at Y.
∠DCA = c/2
∠ACF = ∠FDA = b/2 (same segment)
∠YCX = ∠YDX (same segment); but ∠YDX is ∠FDA, so ∠YCX = b/2 as well
∠EDF = ∠EDC + ∠CDF = a/2 + b/2 (I’m sure you get the idea by now)
So ∠ECF = 180 – a/2 – b/2 (opposite angle of cyclic quadrilateral)
∠FCX = a/2 + b/2 (angles on a straight line)
So ∠FCY = ∠FCX – ∠YCX = a/2
∠DCY = ∠DCA + ∠ACF + ∠FCY = a/2 + b/2 + c/2 = (a + b + c)/2 = 90
So by the converse of Thales’ theorem, DY is a diameter of Λ, meaning that the centre of Λ lies on DY.
Obviously I could’ve gotten another result like this by extending EC in the other direction.