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Sia ? 6 years, 3 months ago
Given: In a {tex}\triangle{/tex}ABC, consider AD be the bisector of {tex}\angle{/tex}A which meets BC in D such that BD = DC.
To Prove: {tex}\triangle{/tex}ABC is an isosceles triangle i.e. AB = AC.
Construction: Produce AD to E such that AD=DE and then join CE.
Proof: In {tex}\Delta ABD \;and \;\Delta ECD,{/tex} we have
BD = CD [Given]
AD = ED [By construction]
and {tex}\angle{/tex}ADB = {tex}\angle{/tex}EDC [Vertically opposite angles]
Therefore, {tex}\Delta ABD=\Delta ECD{/tex} [SAS congruence criterion]
So, AB = EC [Corresponding parts of congruent triangles are congruent] ---- (i)
and {tex}\angle{/tex}BAD = {tex}\angle{/tex}CED [Corresponding parts of congruent triangles are congruent] ---- (ii)
Also, {tex}\angle{/tex}BAD = {tex}\angle{/tex}CAD [Given] ---- (iii)
Therefore, From (ii) and (iii) we get
{tex}\angle{/tex}CAD = {tex}\angle{/tex}CED
So, AC = EC [Sides opposite to angles are equal]....... (iv)
From (i) and (iv), we get
AB = AC
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC is an isosceles triangle.
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Harshit Garg 6 years, 3 months ago
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