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Sia ? 6 years, 4 months ago
{tex}{/tex}Given: {tex}\Delta{/tex}ABC in which MX {tex}\|{/tex} AB and NX{tex}\|{/tex}AC. MN meets BC produced at T.
To prove: TX2 = TB {tex}\times{/tex} TC
Proof: In {tex}\Delta MTX ,{/tex}
{tex} X M \| B N{/tex} {given}
{tex} \therefore \quad \frac { T B } { T X } = \frac { T N } { T M }{/tex} ...............(i) [by basic proportionality theorem]
In {tex} \Delta{/tex} TMC, we have
{tex} X N \| C M{/tex}{ given}
{tex} \therefore \quad \frac { T X } { T C } = \frac { T N } { T M }{/tex} ..............(ii) [by basic proportionality theorem]
From equations (i) and (ii), we get,
{tex}\Rightarrow{/tex}{tex} \frac { T B } { T X } = \frac { T X } { T C }{/tex}
{tex} \Rightarrow \quad T X ^ { 2 } = T B \times T C{/tex}
Hence proved.
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