In fig., AD is a median of a triangle ABC and AM⊥BC. Prove that: (i) AC²=AD²+BC.DM+(BC/2)² (ii) AB²=AD²-BC.DM+(BC/2)² (iii) AC²+AB²=2AD²+(1/2)BC²

It is given that ∠AMD=90⁰
Referring to the figure, we can say that ∠ADM<90⁰ and ∠ADC>90⁰
Now,
(i)To prove: AC²=AD²+BC.DM+(BC/2)²
In ΔADC, ∠ADC is an obtuse angle.
∴AC²=Ad²+DC²+2DC.DM
⇒AC²=AD²+(BC/2)²+2.(BC/2).DM
⇒AC²=AD²+BC.DM+(BC/2)²
(ii)To prove: AB²=AD²-BC.DM+(BC/2)²
In ΔABD, ∠ADM is an obtuse angle.
∴AB²=AD²+BD²-2;BD.DM
⇒AB²=AD²+(BC/2)²-2.(BC/2).DM
⇒AB²=AD²-BC.DM+(BC/2)²
(iii)To prove: AC²+AB²=2AD²+(1/2)BC²
From the result of (i) and (ii), adding those, we get
AC²+AB²=2AD²+(1/2)BC²