Class 9 Maths Chapter 7 | Triangles | Exercise 7.2 Solutions

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In question 1. just by just proving triangle AOB congruent to triangle AOC by SAS rule we can find both i and ii
AO=AO(common)
angle ABC = angle ACB
therefore, 1/2 ABC = 1/2 ACB
angle ABO = angle ACO
AB = AC(given)
triangle AOB congruent to triangle AOC
i. OB = OC(CPCT)
ii. angle BAO = angle CAO(CPCT)
AO bisects angle A
sir is this correct??

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You cannot provide ∆AOB congruent ∆AOC by S.A.S because in S.A.S rule angle must be included between two sides.

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