Given a particular triangle that has been constructed, I want to prove that one of the angles must be $> 45$ degrees. 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! 
 A: Let's choose some coordinates, w.l.o.g.:
$$X=(0,0)\qquad Y=(1,0)\qquad Z=(t,u)$$
Then you have
$$XB:ux=(t+1)y\qquad YC:tx+uy=t$$
which intersect in
$$P=(t^2+t,tu)/(u^2+t^2+t)$$
Now the (quadratic) condition $\angle XZP=\angle PZY$ can be rephrased as
\begin{align*}
\frac{\langle X-Z,P-Z\rangle}{\lVert X-Z\rVert\cdot\lVert P-Z\rVert} &=
\frac{\langle P-Z,Y-Z\rangle}{\lVert P-Z\rVert\cdot\lVert Y-Z\rVert} \\
\langle X-Z,P-Z\rangle^2\cdot\lVert Y-Z\rVert^2 &=
\langle P-Z,Y-Z\rangle^2\cdot\lVert X-Z\rVert^2 \\
(u^2+t^2-t)^2(u^2+t^2-2t+1) &=
\left(\frac{u^4+2u^2t^2+t^4-t^3-tu^2-t^2+t}{u^2+t^2+t}\right)^2(t^2+u^2) \\
\end{align*}
$$u^2  (t^6 + 3t^4u^2 + 3t^2u^4 + u^6 - 2t^5 - 4t^3u^2 - 2tu^4 + 2t^3 - t^2)=0$$
So if $u\neq 0$, then the second term characterizes all possible locations of the third corner. That's an algebraic curve of degree $6$.
You have to show that for $0<t<1$ this only results in real solutions which lie within the circle whose inscribed angle would be $45°$. Next I'd plot that algebraic curve:

As you can see, the blue curve stays within the red circles, so the angle is larger than $45°$. And the curve stays outside the orange circle, so the triangle is acute in all these cases.
One possible triangle satisfying your requirements would be the following example:

