Ex 7.2, 3

Transcript

Ex7.2, 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. Given: Given ∆ABC is isosceles with AB & AC equal, i.e. AB = AC BE and CF are altitudes. So, ∠AEB = 90° and ∠AFC = 90° To prove: BE = CF Proof: We take ΔABE and ΔACF, In ΔABE and ΔACF, ∠AEB = ∠AFC ∠A = ∠A AB = AC ∴ ΔABE ≅ ΔACF ⇒ BE = CF Hence proved