1. Given: Triangle $ABC$ with point $P$ inside such that $\angle PAC = \angle PBC$. Perpendiculars from $P$ to $BC$ and $CA$ meet these lines at $L$ and $M$ respectively. $D$ is the midpoint of $AB$.

2. Objective: Prove that $DL = DM$.

3. Step 1: Consider the midpoints of lines $BP$ and $AP$, denoted as $E$ and $F$ respectively. Since $D$ is the midpoint of $AB$, we have:
$$D = \text{midpoint of } AB$$
$$E = \text{midpoint of } BP$$
$$F = \text{midpoint of } AP$$

4. Step 2: Since $D$ is the midpoint of $AB$, $DE$ and $DF$ are medians of triangles $DBP$ and $DAP$ respectively. By the midpoint theorem, $DE$ and $DF$ are parallel to $AP$ and $BP$ respectively and half their lengths.

5. Step 3: Since $\angle PAC = \angle PBC$, triangles $PAC$ and $PBC$ are similar by AA similarity criterion. This implies that the perpendiculars from $P$ to $BC$ and $CA$ (i.e., $PL$ and $PM$) are equal in length.

6. Step 4: Since $PL \perp BC$ and $PM \perp CA$, $PL$ and $PM$ are the altitudes from $P$ to $BC$ and $CA$ respectively. Therefore, $L$ and $M$ are the feet of the perpendiculars from $P$ to $BC$ and $CA$.

7. Step 5: By the properties of perpendiculars and midpoints, we have:
$$DF = EL \quad \text{and} \quad DE = FM$$
This is because $DF$ and $EL$ are both half the length of $AP$ and $DE$ and $FM$ are both half the length of $BP$.

8. Step 6: By angle chasing, we can show that $\angle DFM = \angle DEL$. Since $DF = EL$ and $DE = FM$, triangles $DFM$ and $DEL$ are congruent by the SAS (Side-Angle-Side) criterion.

9. Conclusion: Since $\triangle DFM \cong \triangle DEL$, it follows that $DL = DM$.

$\blacksquare$