In triangle ABC, D is the midpoint of AC such that BD = 1/2 AC show that angle ABC is a right angle

In ΔADB , AD = BD

⇒∠DAB = ∠DBA = ∠x  (angles opposite equal sides)

Similarly in ΔDCB , BD=CD

⇒∠DBC = ∠DCB = ∠y

In ΔABC , by angle sum property

∠ABC + ∠BCA +∠CAB = 180º

⇒∠x + ∠x + ∠y + ∠y = 180º

⇒ 2(∠x + ∠y) =180º

⇒ ∠x + ∠y = 90º

⇒ ∠ABC = 90º

hence ABC is right angled