Got this proof from the internet.
As before, ABC is the triangle, BY, CZ the bisectors of B, C, with Y on AC, Z on AB. We have BY = CZ. Let E be the intersection of BY and CZ. Connect A, E.
Find the point D outside of edge AB so that BD = AZ and DY = AC.
So triangle DBY is congruent with AZC. Let the bisector of
A, D, B, Y are concyclic (
Then A, D, F, E are concyclic (
Since AE = DF (DBY congruent with AZC) , we have AD // EF.
Therefore,
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