Let $S$ be the point such that $\overline{AS}$, $\overline{OI}$, $\overline{BC}$ concur. Let $M = \overline{AI} \cap (ABC)$ and $T = \overline{MD} \cap (ABC)$. From this, we have $DSHT$ cyclic and $S$, $D$, $P$ collinear. Since $\overline{HD} \parallel \overline{NM}$, we have $H$, $N$, $S$ collinear by Reim's. Furthermore, if $P' = \overline{PI} \cap (ABC)$, then $$\angle NP'P = 90^\circ - \angle PAM = 90^\circ - \angle PAI = \tfrac 12 \angle AIP = \angle P'IO,$$so $\overline{NP'} \parallel \overline{IO}$. Therefore $$\angle IPD = \angle P'PS = \angle P'NS = \angle P'NX = \angle NXO = \angle HXI,$$implying the conclusion.

Thanks to MP8148 after seeing 2020 Taiwan TST Round 2 Mock Exam P6 (solution uses this), I solved in a not so different way. If $D'$ is the reflection of $D$ across $EF$. $TDHD' \sim TPNA \implies$ spiral similarity at $T$ sends $DH$ to $NP \implies NH$ and $PD$ concur on $(ABC)$. $$\angle NSP= \angle NAP=90-\angle LAI=\angle LIP \implies \angle XID=\angle HIL=\angle HIP- \angle HSP=\angle SHI +\angle SPI$$as desired. $\textbf{Note:}$ Proof of $TDHD' \sim TPNA$ can be done by noticing $TKD \sim TLP$ and $TAN \sim TD'H$ where $K$ is the midpoint of $TD'$.

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Can you show me the link of 2020 Taiwan TST R2 Mock exam P6. I cannot find it.Tks!

SerdarBozdag wrote: Thanks to MP8148 after seeing 2020 Taiwan TST Round 2 Mock Exam P6 (solution uses this), I solved in a not so different way. If $D'$ is the reflection of $D$ across $EF$. $TDHD' \sim TPNA \implies$ spiral similarity at $T$ sends $DH$ to $NP \implies NH$ and $PD$ concur on $(ABC)$. $$\angle NSP= \angle NAP=90-\angle LAI=\angle LIP \implies \angle XID=\angle HIL=\angle HIP- \angle HSP=\angle SHI +\angle SPI$$as desired. $\textbf{Note:}$ Proof of $TDHD' \sim TPNA$ can be done by noticing $TKD \sim TLP$ and $TAN \sim TD'H$ where $K$ is the midpoint of $TD'$. Can you show me the link of 2020 Taiwan TST R2 Mock exam P6. I cannot find it.Tks!

https://artofproblemsolving.com/community/c6h2073612p14866139