1. **Stating the problem:** We are given that triangles ABC and EHK are congruent ($\triangle ABC \equiv \triangle EHK$), and segments MBHP and NCHL are drawn such that $BC \cong BH$. We know $m\angle NCD = 62^\circ$ and need to find $m\angle PHK$. 2. **Understanding the congruence:** Since $\triangle ABC \equiv \triangle EHK$, corresponding angles and sides are equal: - $\angle ABC = \angle EHK$ - $BC = HK$ - $AB = EH$ - $AC = EK$ 3. **Given $BC \cong BH$:** This means $BH = BC = HK$ (since $HK$ corresponds to $BC$ in the congruence). 4. **Analyzing angles:** $m\angle NCD = 62^\circ$ is given. Since $NCHL$ and $MBHP$ are drawn with points on the lines intersecting at $H$, and $BC \cong BH$, the angles around $H$ relate to those in the congruent triangles. 5. **Using vertical angles and congruence:** Because $\triangle ABC \equiv \triangle EHK$, $m\angle ABC = m\angle EHK$. Since $m\angle NCD = 62^\circ$ corresponds to $m\angle PHK$ by the given configuration and congruence, we conclude: $$m\angle PHK = 62^\circ$$ **Final answer:** $$m\angle PHK = 62^\circ$$