Question

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X is a point on the side BC of a ∆ABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in M and AC in M. MN produced at T. Prove that TX² = TB×TC.


➡ CLASS - X
➡ CH -6
➡ TRIANGLES​

Answer 1

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Here is your answer

Here given that,

X is a point on the side BC of a ∆ABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in M and AC in M.

MN produced at T.

So, We have to prove that

TX² = TB×TC.

⇒ TX / TC = XN / CM = TN / TM

⇒ TX × TM = TC × TN ....(i)

Again,

∆TBN ~ ∆TXM

⇒ TB / TX = BN / XM = TN / TM

⇒ TM = (TN × TX) / TB ... (ii)

Using (ii) in (i),

We get,

TX2 × TN/TB = TC × TN

⇒ TX2 = TC × TB

Hope its help you ✌✌

Answer 2

So this is the answer you need...

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