Ex 7.2, 4 - ABC is a triangle in which altitudes BE & CF - Opposite Angles of equal sides

  1. Class 9
  2. Important Questions for Exam - Class 9
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Ex7.2, 4 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that (i)  ∆ABE ≅  ∆ACF (ii) AB = AC, i.e., ABC is an isosceles triangle. Comparison with Ex 7.2 , 3 In Ex 7.2 , 3 AB = AC given , we have to prove BE = CF In Ex 7.2 , 4 Here, BE = CF given , we have to prove AB = AC Ex7.2, 4 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that (i)  ∆ABE ≅  ∆ACF (ii) AB = AC, i.e., ABC is an isosceles triangle. Given: BE = CF BE and CF are altitudes. So, ∠AEB = 90° and ∠AFC = 90° To prove: ΔABE ≅ ΔACF & AB = AC Proof: In ΔABE and ΔACF, ∠AEB = ∠AFC ∠A = ∠A BE = CF ∴ ΔABE ≅ ΔACF ⇒ AB = AC Hence proved

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