Question

ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that: AD = BE

Answer 1

In ∆ ABC
AB = BC = CA
We know that
AD is perpendicular to BC and BE is perpendicular to AC
In ∆ ADC and ∆ BEC
∠ADC = ∠BEC = ${90}^{\circ }$
∠ACD = ∠BCE is common
AC = BC are the sides of an equilateral triangle
∆ ADC ≅ ∆ BEC (AAS Axiom)
So, AD = BE (C.P.C.T)