Question 3:
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC
and AB respectively (see the given figure). Show that these altitudes are equal.
Answer:
In ∆AEB and ∆AFC,
∠AEB and ∠AFC (Each 90º)
∠A = ∠A (Common angle)
AB = AC (Given)
∆AEB ≅ ∆AFC (By AAS congruence rule)
BE = CF (By CPCT)
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