Question -
Answer -
Given : In ∆ABC, D is mid-point of BC and DE ⊥ AB, DF ⊥ AC and DE = DF
To Prove : ∆ABC is an isosceles triangle
Proof : In right ∆BDE and ∆CDF,
Side DE = DF
Hyp. BD = CD
∴ ∆BDE ≅ ∆CDF (RHS axiom)
∴ ∠B = ∠C (c.p.c.t.)
Now in ∆ABC,
∠B = ∠C (Prove)
∴ AC = AB (Sides opposite to equal angles)
∴ AABC is an isosceles triangle