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  • X is a point on the side BC of  . XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MN produced meets CB produced at

X is a point on the side BC of  \Delta ABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MN produced meets CB produced at T. Prove that TX^2 =TB \times TC

 

Answers (1)

Here 

XM|| AB , XN || AC \\\\ BN || XM

For \Delta T XM we have BN || XM

BY proportionality theorem 

\frac{TB}{TX} = \frac{TN }{TM}----(1) \\\\ XN || AC \\\\ In\: \: \Delta TMC

we have XN || CM

Again, using proportionality theorem 

\frac{TX }{TC } = \frac{TN }{TM} ---- (2) \\\\

NOw comparing eq (1) and (2) 

\frac{TB }{TX } = \frac{TX }{TC} ---- (2\\\\ TX^2 = TB \times TC

Hence proved  

Posted by

Safeer PP

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