On this page :https://www.cut-the-knot.org/triangle/Morley/MorleyZsolt.shtml
I read
"
There are two special cases that should be mentioned. There is the case a = b = c = 20, when the theorem is obvious. The other case is when two of a, b, c add up to 40 without being equal. In this case, two of the six points coincide but the other four are distinct so the triangle ABC can still be constructed from them.
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There is a mistake : the other case is when a or b or c=30, and in fact we cannot define the line joining two points which are the same...
And I don't see why the case a=b=c=20 is a special case...
So in my opinion this proof is wrong for a triangle ABC with a right angle.