Question

ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.

Answer 1

Let DE and DF be the perpendiculars from D on AB and AC respectively. 

In Δs BDE and CDF, DE = DF     (Given) 

∠BED = ∠CFD = 90° 

∠BD = DC        (∵ D is the mid-point of BC) 

∴ ΔBDE ≅ CDF    (RHS)

⇒ ∠B =∠C         (cpct)

⇒ AC = AB        (Sides opp. equal ∠s are equal) 

ΔABC is isosceles.