1. **Problem Statement:** In triangle ABC, angle B is 90 degrees and D is the midpoint of BC. Prove that $$AC^2 = AD^2 + 3 \cdot CD^2$$. 2. **Given:** - $$\angle B = 90^\circ$$ (right angle at B) - D is midpoint of BC, so $$BD = DC$$. 3. **To Prove:** $$AC^2 = AD^2 + 3 \cdot CD^2$$. 4. **Approach:** Use coordinate geometry or the Pythagorean theorem and midpoint properties. 5. **Assign coordinates:** - Let B be at origin $$B(0,0)$$. - Since $$\angle B = 90^\circ$$, place $$AB$$ along x-axis and $$BC$$ along y-axis. - Let $$AB = c$$ and $$BC = a$$. - So, $$A(c,0)$$ and $$C(0,a)$$. 6. **Find coordinates of D:** - D is midpoint of BC, so $$D = \left(\frac{0+0}{2}, \frac{0+a}{2}\right) = (0, \frac{a}{2})$$. 7. **Calculate lengths:** - $$AC^2 = (c-0)^2 + (0 - a)^2 = c^2 + a^2$$. - $$AD^2 = (c - 0)^2 + \left(0 - \frac{a}{2}\right)^2 = c^2 + \frac{a^2}{4}$$. - $$CD^2 = (0 - 0)^2 + \left(a - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2 = \frac{a^2}{4}$$. 8. **Substitute into the expression:** $$AD^2 + 3 \cdot CD^2 = c^2 + \frac{a^2}{4} + 3 \times \frac{a^2}{4} = c^2 + \frac{a^2}{4} + \frac{3a^2}{4} = c^2 + a^2$$. 9. **Conclusion:** $$AC^2 = AD^2 + 3 \cdot CD^2$$ is proved. This completes the proof.