In triangle ABC ,AB=AC,BD PERPENDICULAR AB. …
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Sia ? 5 years, 3 months ago
Given: {tex}\Delta ABC{/tex} with AB = AC
And AD = CD, AE = BE.
To prove: BD = CE
Proof: In {tex}\Delta ABC{/tex} we have
AB = AC [Given]
{tex} \Rightarrow \;\frac{1}{2}AB = \frac{1}{2}AC{/tex}
{tex}\Rightarrow{/tex} AE = AD
[{tex}\because{/tex} D is the mid-point of AC and E is the mid-point of AB]
Now, in {tex}\Delta ABD{/tex} and {tex}\Delta ACE{/tex}, we have
AB = AC [Given]
{tex}\angle A = \angle A{/tex} (Common angle]
AE = AD [Proved above]
SO, by SAS criterion of congruence, we have
{tex}\Delta ABD \cong \Delta ACE{/tex}
{tex}\Rightarrow{/tex} BD = CE [CPCT]
Hence, proved.
0Thank You