In a triangle ABC, D is the mid-point of side AC such that BD = $\frac{1}{2}$AC. Find ∠ABC = ?

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = $\frac{1}{2}$AC …(i)
Also, BD = $\frac{1}{2}$AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]

⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90°