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Converse of Basic Proportionality Theorem

Let X be any ...

Question

Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that $T X^{2} = T B \times T C$ .

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Solution

In $△ T X M$ ,
$X M ∥ B N$
Therefore,
$\frac{T B}{T X} = \frac{T N}{T M}$ ......(1)

In $△ T M C$ ,
$\frac{T X}{T C} = \frac{T N}{T M}$ .......(2)

From 1 and 2, we have
$\frac{T B}{T X} = \frac{T X}{T C}$

$T X^{2} = T B \times T C$

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{T X}^{2} = T B \times T C

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Let X be any point on the side BC of △ ABC . XM and XN are drawn parallel to BA and CA . MN meets CB produced to T . Prove that {TX}^{2} = TB . TC

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