A variant of Morley’s trisector theorem involving external angles

Question

This is a question about a variant of Morley’s trisector theorem.

For any $\triangle ABC$, if the trisection lines of one of the angles and the trisection lines of the other two external angles intersect at points $X$, $Y$, $Z$, then $\triangle XYZ$ is a regular triangle.

Here is what I have tried.

1. Suppose the trisection lines of the external angles $B$ and $C$ intersect at $X$, $Q$, $R$, $S$, as shown. We can get $\angle BSX = \angle CSX$.

2. Draw $\angle SXZ = \angle SXY =30^\circ$. Intersect $BS$ and $CS$ at $Z$ and $Y$. We know $\triangle XYZ$ is a regular triangle.

3. Now I need to prove that $AY$ and $AZ$ trisect $\angle A$. First reflect point X by line BS,CS to X' and X",which X' and X"are on AB and AC.We have x'z=xz=xy=x"y.

4.Now I want to prove A,X',Z,Y,X" are on the same circle.Stucking here. I am trying to use the same way Dr.Naraniengar proof this question.

Any help is appreciated! Thank you very much.:)

Answer 1

I've work out the problem!Yay! :)

Suppose the trisection lines of the external angles B and C intersect at X, Q, R, S, as shown. We can get ∠BSX=∠CSX.

Draw ∠SXZ=∠SXY=30∘. Intersect BS and CS at Z and Y. We know △XYZ is a regular triangle.

Now I need to prove that AY and AZ trisect ∠A. First reflect point X by line BS,CS to X' and X",which X' and X"are on AB and AC.We have X'Z=XZ=XY=X"Y.

Now I want to prove A,X',Z,Y,X" are on the same circle.

Draw the circumcircle of X',Z,Y,X" ,which is circle O. We know ∠BSX=∠CSX=90-β-γ, so ∠BZX=30+90-β-γ=120-β-γ ,

∠X'ZY=∠X"YZ=300-2β-2γ,

∠X'ZY+∠X"YZ+OX'Z+OX"Y=900-6β-6γ,

∠X'OX"=540-(900-6β-6γ)=6β+6γ-360,

The inscribed angle of ∠X'OX" is 3β+3γ-180,which is equal to ∠A=180-(180-3β)-(180-3γ)=3β+3γ-180, hence A is on circle O.

since X'Z=ZY=X"Y,finally got ∠X'AZ=∠ZAY=∠YAX".