1. **Problem statement:** In triangle ABC, angle B is 90 degrees, and T is the midpoint of BC. Prove that $$AC^2 = AD^2 + 3CD^2$$ using the Pythagorean theorem. 2. **Setup and notation:** Since B is the right angle, triangle ABC is right-angled at B. 3. **Key points:** T is the midpoint of BC, so $$BT = TC$$. 4. **Using Pythagoras theorem:** In right triangle ABC, $$AC^2 = AB^2 + BC^2$$. 5. **Expressing lengths:** Let’s denote: - $$BC = 2x$$ (since T is midpoint, BC is twice BT or TC) - $$BT = TC = x$$ 6. **Point D:** The problem mentions AD and CD, but D is not defined in the problem statement. Assuming D is point T (midpoint of BC), then D = T. 7. **Calculate AD and CD:** - Since D is midpoint of BC, CD = x - Triangle ABD is right-angled at B (since B is 90 degrees), so by Pythagoras: $$AD^2 = AB^2 + BD^2$$ 8. **Relate BD and CD:** Since D is midpoint, $$BD = x$$ and $$CD = x$$. 9. **Rewrite AC^2:** $$AC^2 = AB^2 + BC^2 = AB^2 + (2x)^2 = AB^2 + 4x^2$$ 10. **Rewrite AD^2 + 3CD^2:** $$AD^2 + 3CD^2 = (AB^2 + BD^2) + 3x^2 = AB^2 + x^2 + 3x^2 = AB^2 + 4x^2$$ 11. **Conclusion:** Since both expressions equal $$AB^2 + 4x^2$$, we have $$AC^2 = AD^2 + 3CD^2$$ Thus, the statement is proved using the Pythagorean theorem.