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Pythagoras Theorem

Let X be any ...

Question

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

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Solution

$In △ TXM, seg BN ∥ seg XM . Therefore, according to the basic proportionality theorem, \frac{TN}{NM} = \frac{TB}{BX} . . . (1) S eg XN ∥ S eg CM Therefore, according to the basic proportionalit y theorem, \frac{TN}{NM} = \frac{TX}{CX} . . . (2) \Rightarrow \frac{TB}{BX} = \frac{TX}{CX} [From (1) & (2)] \Rightarrow \frac{BX}{TB} = \frac{CX}{TX} (Invertendo) \Rightarrow \frac{BX + TB}{TB} = \frac{CX + TX}{TX} (Componendo) \Rightarrow \frac{TX}{TB} = \frac{TC}{TX} \Rightarrow {TX}^{2} = TB . TC Hence proved .$

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

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