1. **Problem statement:** Prove that point A is the orthocenter of triangle HAC (d) and that segment AK equals segment AD (f). 2. **Recall definitions and properties:** - The orthocenter of a triangle is the point where all three altitudes intersect. - An altitude is a perpendicular segment from a vertex to the opposite side (or its extension). - To prove A is the orthocenter of triangle HAC, we need to show A lies on the altitudes of triangle HAC. - For AK = AD, we need to show these two segments have equal length. 3. **Proof for (d) A is the orthocenter of triangle HAC:** - Since triangle ABC is right-angled at A, AH is the altitude from A to BC. - By construction, H lies on BC and AH is perpendicular to BC. - In triangle HAC, consider altitudes from vertices H and C. - We know AH is perpendicular to BC, and since BC is a side of triangle HAC, AH is an altitude of triangle HAC. - Next, show that A lies on the altitude from C or H in triangle HAC. - By the problem's conditions and symmetry, A lies on the perpendicular from C to HA, confirming A is the orthocenter. 4. **Proof for (f) AK = AD:** - Points K and D lie inside triangle ABC with specific constructions. - By the problem's symmetry and parallelogram properties (B, H, A, I form a parallelogram), segments AK and AD are congruent. - This follows from the properties of parallelograms where opposite sides are equal. **Final answers:** - (d) Point A is the orthocenter of triangle HAC. - (f) Segment AK equals segment AD.