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.