In ∆ABC ,angle B= angle C …
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Sia ? 6 years, 4 months ago
In {tex} \Delta{/tex}ABC, we have
{tex} \angle{/tex}B = {tex} \angle{/tex}C
{tex} \Rightarrow{/tex} AC = AB [Sides opposite to equal angles are equal]
{tex} \Rightarrow{/tex} AE + EC = AD + DB
{tex} \Rightarrow{/tex} AE + CE = AD + BD
{tex}\Rightarrow{/tex} AE + CE = AD + CE [{tex}\because{/tex} BD = CE]
{tex}\Rightarrow{/tex} AE = AD
Thus, we have
AD = AE and BD = CE
{tex}\therefore \quad \frac { A D } { B D } = \frac { A E } { C E }{/tex}
{tex}\Rightarrow \quad \frac { A D } { D B } = \frac { A E } { E C }{/tex}
Therefore,by the converse of basic proportionality theorem, we have,
{tex} \quad D E \| B C{/tex}
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