ABC is a triangle in which …
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Sia ? 6 years, 5 months ago
Draw {tex}A E \perp B C{/tex}
In {tex}\Delta{/tex}AEB and {tex}\Delta{/tex}AEC, we have
AB = AC,
AE = AE [Common]
and, {tex}\angle{/tex}B = {tex}\angle{/tex}C [{tex}\because{/tex} AB = AC]
{tex}\therefore \quad \Delta A E B \cong \Delta A E C{/tex}
{tex}\Rightarrow{/tex} BE = CE [by CPCT]
Since {tex}\Delta{/tex}AED and {tex}\Delta{/tex}ABE are right triangles right-angled at E.
Therefore,
{tex}\Rightarrow{/tex} AD2 = AE2 + DE2 and AB2 = AE2 + BE2
{tex}\Rightarrow{/tex} AB2 - AD2 = BE2 - DE2
{tex}\Rightarrow{/tex} AB2 - AD2 = (BE+ DE) (BE - DE)
{tex}\Rightarrow{/tex} AB2 - AD2 = (CE+ DE) (BE - DE) [{tex}\because{/tex} BE = CE ]
{tex}\Rightarrow{/tex} AB2- AD2 = CD·BD
Hence, AB2 - AD2 = BD·CD
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