i) Apollonius Theorem:
"In a triangle ABC, if AD is the median to BC, then AB² + AC² = 2(AD² + BD²)"
ii) As I am not sure of your knowledge on Cosine Law of triangle, let me provide the solution using basic geometrical concepts applied in high school geometry.
iii) Draw an altitude from vertex A to BC, to meet BC at E. [E may be within B & D or withing D & C, depends upon AB < AC or AB > AC; let us here take AB < AC, so E is between B & D]
iv) Applying Pythagoras theorem to respective right triangles,
From right triangle, AEB, AB² = AE² + BE² ------ (1)
From right triangle, AED, AD² = AE² + ED² ------ (2)
(1) - (2): AB² - AD² = BE² - ED² = (BD - DE)² - ED² = BD² - 2BD*DE + ED² - ED²
==> AB² - AD² = BD² - 2BD*DE ------- (3)
v) Similarly considering the right triangles, AEC & AED, we get,
AC² - AD² = DE² - ED² = (CD + DE)² - DE² = CD² + 2CD*DE + DE² - DE²
==> AC² - AD² = CD² + 2CD*DE = BD² + 2BD*DE ------ (4)
[Since AD is the median, D is mid point of BC; hence BD = DC]
vi) Adding (3) & (4)
==> AB² + AC² - 2AD² = 2BD²
==> AB² + AC² = 2(AD² + BD²) [Proved]