Question

ABC is an isosceles triangle in which altitudes BE & CF are drawn
to sides Ac and AB respectively. Show that these altitudes are
equal
​

Answer 1

ΔABC is an isosceles triangle.

          ∴ AB = AC

          ⇒ ∠ACB = ∠ABC

[∴ Angles opposite to equal sides are equal]

          Now, in ΔBEC and ΔCFB, we have

          ∠EBC = ∠FCB

[Proved]

          BC = CB

[Common]

          and ∠BEC = ∠CFB

[Each = 90°]

          ∴ ΔBEC ≌ ΔCFB

[Using ASA criteria]

          ⇒ Their corresponding parts are equal.

          i.e. BE = CF

hope it helps you