Question:

<p><strong>Step 1:</strong> Understand the reflection property. Since D is the reflection of A with respect to the perpendicular bisector of BC, we have DB = AB and DC = AC.</p><p><strong>Step 2:</strong> Recognize that D lies on a circle. Because D is the reflection of A across the perpendicular bisector of BC, the point D is positioned such that △DBC is congruent to a specific configuration. In fact, DB = AB and DC = AC.</p><p><strong>Step 3:</strong> Apply the property of the orthocenter. Since A is the orthocenter of △ABC, we have: BA ⊥ AC (or more precisely, the altitude from B to AC passes through A, etc.). By the reflection property, when we reflect A to get D across the perpendicular bisector of BC, the angles transform such that D lies on the circle with BC as chord.</p><p><strong>Step 4:</strong> Find the orthocenter of △DBC. The crucial insight is that since A was the orthocenter of △ABC with altitudes meeting at A, and D is the reflection of A across the perpendicular bisector of BC, the orthocenter of △DBC is the reflection of A across BC's perpendicular bisector, which is exactly D itself. This is because the altitude from D in △DBC passes through specific points that make D the orthocenter.</p><p><strong>Step 5:</strong> Verify using angle properties. The altitudes of △DBC must meet at a point. By the reflection symmetry and the property that reflections preserve perpendicularity relationships across the perpendicular bisector, the point D itself serves as the orthocenter of △DBC.</p><p><strong>∴ Answer:</strong> A</p>