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.

Advertisement Remove all ads

#### Solution

In ΔAEB and ΔAFC,

∠AEB and ∠AFC (Each 90º)

∠A = ∠A (Common angle)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)

⇒ BE = CF (By CPCT)

Concept: Properties of a Triangle

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads