Problem 1650

1. Given Conditions and Initial Setup:
- Point $P$ is inside triangle $ABC$ such that $\angle PAC = \angle PCB$.
- Projections of $P$ onto $BC$, $CA$, and $AB$ are $X$, $Y$, and $Z$ respectively.
- $O$ is the circumcenter of $\triangle XYZ$.
- $H$ is the foot of the altitude from $B$ to $AC$.
- $N$ is the midpoint of $AC$.
- $T$ is the point such that $TYPO$ is a parallelogram.

2. Angle Chasing and Initial Observations:
- Since $\angle PAC = \angle PCB$, point $P$ lies on the angle bisector of $\angle ACB$.
- Let the line through $P$ and $X$ parallel to $AC$ meet $BH$ and $AB$ at $H'$ and $A'$ respectively.

3. Using the Angle Equality Condition:
- By the given condition, $\angle ZYX = \angle A$.
- Therefore, $\angle ZOX = 2 \angle BA'X$, implying $A'$ lies on the circumcircle of $\triangle XYZ$.

4. Collinearity and Angle Chasing:
- Since $H'$ lies on the circle $(BXPZ)$ because $\angle BH'P = 90^\circ$, we have:
$$\angle A'XH' = \angle XH'P = 90^\circ - \angle BZX = \angle A'XO$$
- This implies $X, O, H'$ are collinear.

5. Similarity of Triangles:
- We need to show $\triangle THN \sim \triangle PBC$.
- Let the pedal feet from $O$ to $AC$ be $L$.
- We have $\triangle THL \sim \triangle PBX$.

6. Homothety and Ratio Analysis:
- To prove $\frac{HL}{LN} = \frac{BX}{XC}$, consider a homothety centered at $B$ mapping $X, A', O$ to $C, A, O'$.
- Since $CO' = AO'$, we have $NO' \parallel OL \parallel BH$.
- Therefore, $\frac{HL}{LN} = \frac{BO}{OO'} = \frac{BX}{XC}$.

7. Conclusion:
- Since all the conditions and ratios are satisfied, we conclude that $\triangle THN \sim \triangle PBC$.

$\blacksquare$