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Sia ? 5 years, 9 months ago
Given: In a △ABC, consider AD be the bisector of ∠A which meets BC in D such that BD = DC.
To Prove: △ABC is an isosceles triangle i.e. AB = AC.
Construction: Produce AD to E such that AD=DE and then join CE.
Proof: In ΔABDandΔECD, we have
BD = CD [Given]
AD = ED [By construction]
and ∠ADB = ∠EDC [Vertically opposite angles]
Therefore, ΔABD=ΔECD [SAS congruence criterion]
So, AB = EC [Corresponding parts of congruent triangles are congruent] ---- (i)
and ∠BAD = ∠CED [Corresponding parts of congruent triangles are congruent] ---- (ii)
Also, ∠BAD = ∠CAD [Given] ---- (iii)
Therefore, From (ii) and (iii) we get
∠CAD = ∠CED
So, AC = EC [Sides opposite to angles are equal]....... (iv)
From (i) and (iv), we get
AB = AC
∴ △ABC is an isosceles triangle.
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Harshit Garg 5 years, 9 months ago
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