1. **Problem Statement:** Given quadrilateral ABCD with diagonal AC, points E and F lie on AC such that E is closer to D and F is between A and E. We know lengths: $BD=12$, $AE=8$, $CF=5$, $DE=7$, $BF=5$, and $AC=13$. Right angles at E and F indicate perpendiculars from B and D to AC. Find the lengths or relationships as needed. 2. **Understanding the Setup:** Since E and F lie on AC, and BF and DE are perpendiculars to AC, triangles BFA and DCE are right triangles. 3. **Using the Pythagorean theorem:** - For triangle BFA, with right angle at F: $$BF^2 + AF^2 = AB^2$$ - For triangle DCE, with right angle at E: $$DE^2 + CE^2 = DC^2$$ 4. **Expressing segments on AC:** Since $AC=13$, and points F and E lie on AC with $AF + FE + EC = 13$. Given $AE=8$, so $AF + FE = 8$. Given $CF=5$, so $CF = AC - AF = 13 - AF = 5$ which implies $AF = 8$. 5. **Calculate AB:** From triangle BFA: $$BF=5, AF=8$$ $$AB = \sqrt{5^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89}$$ 6. **Calculate DC:** From triangle DCE: Given $DE=7$, and $CE = AC - AE = 13 - 8 = 5$ $$DC = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74}$$ 7. **Calculate BD:** Given $BD=12$ (already provided). **Final answers:** $$AB = \sqrt{89} \approx 9.43$$ $$DC = \sqrt{74} \approx 8.60$$ $$BD = 12$$