Question
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e., ∆ABC is an isosceles triangle.
Solution
Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal.
To Prove: (i) ∆ABE = ∆ACF
(ii) AB = AC, i.e., ∆ABC is an isosceles triangle.
Proof: (i) In ∆ABE and ∆ACF
BE = CF | Given
∠BAE = ∠CAF | Common
∠AEB = ∠AFC | Each = 90°
∴ ∆ABE ≅ ∆ACF | By AAS Rule
(ii) ∆ABE ≅ ∆ACF | Proved in (i) above
∴ AB = AC | C.P.C.T.
∴ ∆ABC is an isosceles triangle.
