Question

In a ∆ABC, D is the mid-point of AC such that BD = $\frac{1}{2}$ AC. Show that ∠ABC is a right angle.

Answer 1

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

D is the mid-point of AC.

∴ AD = CD = $\frac{1}{2}$AC

⇒ AD = CD = BD (BD = $\frac{1}{2}$AC)

In ∆ABD,

AD = BD

∴ ∠ABD = ∠A .....(1) (In a triangle, equal sides have equal angles opposite to them)

In ∆CBD,

CD = BD

∴ ∠CBD = ∠C .....(2) (In a triangle, equal sides have equal angles opposite to them)

Adding (1) and (2), we get

∠ABD + ∠CBD = ∠A + ∠C

⇒ ∠B = ∠A + ∠C .....(3)

In ∆ABC,

∠A + ∠B + ∠C = 180º (Angle sum property of triangle)

⇒ ∠B + ∠B = 180º [Using (3)]

⇒ 2∠B = 180º

⇒ ∠B = $\frac{180°}{2}$ = 90º

Thus, ∠ABC is a right angle.