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Asked: 2021-08-10T17:40:20+05:30 In:

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects A. Q.2

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Find the simplest solution of ncert class 9th of chapter triangles . Please help me to solve the best solution of exercise 7.3 question number 2 . How i solve this question in a simplest and easiest way please help me to find out this question. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects A.

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    1. Guru

      It is given that AD is an altitude and AB = AC. The diagram is as follows:

      Ncert solutions class 9 chapter 7-18

      (i) In ΔABD and ΔACD,

      ADB = ADC = 90°

      AB = AC (It is given in the question)

      AD = AD (Common arm)

      ∴ ΔABD ΔACD by RHS congruence condition.

      Now, by the rule of CPCT,

      BD = CD.

      So, AD bisects BC

      (ii) Again, by the rule of CPCT, BAD = CAD

      Hence, AD bisects A.

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