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Question
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
- ΔABE ≅ ΔACF
- AB = AC, i.e., ABC is an isosceles triangle.
Solution
i. In △ABE and △ACF, we have
∠AEB = ∠AFC ...[Each = 90° as BE ⊥ AC and CF ⊥ AB]
∠A = ∠A ...[Common]
BE = CF ...[Given]
∴ △ABE ≌ △ACF ...[By AAS congruence rule]
ii. Since, △ABE ≌ △ACF
∴ AB = AC ...[By Corresponding parts of congruent triangles]
⇒ ABC is an isosceles triangle.
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