prove that the bisector of tbe …
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Yogita Ingle 6 years ago
∠B = ∠C
Sides AB = AC.
The AO line is a angular bisect.
=> ∠BAO = ∠OAC = ∠A/2
In ΔABC, ∠A + ∠B + ∠C = 180
Since ∠B = ∠C,
∠A + 2∠B = 180
In ΔBAO, ∠B + ∠AOB + ∠OAB = 180
Since ∠OAB = ∠A/2
∠B + ∠AOB + ∠A/2 = 180
2∠B + 2∠AOB + ∠A = 360 (Multiply both sides with 2)
From above ∠A + 2∠B = 180
Hence 180 + 2∠AOB = 360
∠AOB = 90.
Similar way, we can prove that ∠AOC = 90.
Now in ΔAOB, ΔAOC,
AB = BC, AO = AO and all angles are same.
That means ΔAOB and ΔAOC are congruent.
Therefore BO = CO.
That implies, O is center of BC. AO bisects BC.
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