ABC is a triangle in which AB = AC and D is any point on BC. Which of the following statements is/are correct? I. AB² − AD² = CD × BD II. BC² + BD² − DC² = 2BD × BC

In isosceles triangle ABC with AB = AC, drop perpendicular from A to BC meeting BC at M (midpoint). For any point D on BC, by Stewart's theorem or direct computation: AB² − AD² = BD·DC (this is a standard property of isosceles triangle, valid because AM² + MD² = AD², and AB² = AM² + BM². So AB² − AD² = BM² − MD² = (BM − MD)(BM + MD) = BD·CD when D is between B and M, since BM + MD = BD... actually BM + MD = BD if D between B and M reversed; careful: BM − MD = BD (if D is between M and B), BM + MD = BC/2 + MD... Actually the identity AB² − AD² = BD·DC is well-known for isosceles triangles. So I is correct. For II: BC² + BD² − DC² = BC² + (BD − DC)(BD + DC) = BC² + (BD − DC)·BC = BC(BC + BD − DC) = BC(BD + DC + BD − DC) = BC·2BD = 2BD·BC. So II is also correct. Both I and II are correct.