1. We are given a triangle ABC with a right angle at H, where H is the projection of A on BC, and AH is perpendicular to BC. 2. Given lengths: AB = 6, BH = 4, HC = 5. 3. Since BH + HC = BC, we find BC = 4 + 5 = 9. 4. The right angle at H indicates AH \perp BC, so triangle ABH and AHC are right-angled triangles. 5. Using Pythagoras on triangle ABH: $$AB^2 = AH^2 + BH^2\Rightarrow 6^2 = AH^2 + 4^2\Rightarrow 36 = AH^2 + 16\Rightarrow AH^2 = 20\Rightarrow AH = \sqrt{20}=2\sqrt{5}.$$ 6. Using Pythagoras on triangle AHC: $$AC^2 = AH^2 + HC^2\Rightarrow AC^2 = 20 + 25 = 45\Rightarrow AC = 3\sqrt{5}.$$ 7. Verify coordinate locations: - Point A is given as (-2,1) - Point B is (2,3) 8. Vector \vec{AB} = (2 - (-2), 3 - 1) = (4,2). 9. We can find length AB from coordinates: $$\sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}.$$ This differs from given AB = 6, which suggests coordinates are illustrative but lengths are definitive. 10. Summary: - Length BC = 9 - Length BH = 4, HC = 5 - Length AH = 2\sqrt{5} - Length AB = 6 - Length AC = 3\sqrt{5} This completes the geometric analysis based on given data.