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Suppose you are given an acute triangle $XYZ$ with the following properties:
At $\angle XZY$, the $\angle$ bisector is drawn and extended all the way to $XY$. Lets call the point where it intersects $A$. From $\angle ZXY$ we draw a line to $ZY$ s.t. it cuts $ZY$ into two equal pieces. i.e. the point that intersects $ZY$ is the midpoint call it $B$. And lastly, from $\angle ZYX$ we drop the altitude onto $XZ$. Lets call the point it intersects $C$. Now this triangle was constructed s.t. $ZA \cap XB \cap YC = P$, a point. Suppose we let $\angle XZY = \phi$. How would we prove that $\angle XZY > 45$ degrees for this specific kind of triangle? Obviously my first thought was to suppose that it was $\leq 45$ degrees, and then break it up into two cases. Even then I am unsure. The help would be appreciated!