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Question:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that ΔABE ≅ ΔACF

Solution:

In ΔABE and ΔACF, we have
∠AEB = ∠AFC [Since Each = 90°]
∠BAE = ∠CAF [Common]
and AB = AC [Given]
∴By AAS criterion of congruence, we have ΔABE ≅ ΔACF