Question

X is a point on the side bc of  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 tx2 = tb  tc

Answer 1

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Secondary SchoolMath 5 points

X is a point on the side BC of ∆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×TC...

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Answers

dhruv15819

dhruv15819 Genius

Hey mate

here's the proof

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4.1

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svetlanaoberoy861

Clear nii smj aa ra meko

svetlanaoberoy861

Yah plz

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Shaizakincsem

Shaizakincsem Ace

Thank you for asking this question. Here is your answer:

XM || AB, XN || AC

TX² = TB x TC

BN || XM

For ΔTXM we have

BN || XM

Now we will use BASIC PROPORTIONALITY THEOREM

TB/TX = TN/TM --- (this is equation 1)

XN || AC

XN || CM

In ΔTMC we have

XN || CM

Again we will use BASIC PROPORTIONALITY THEOREM

TX/TC = TN/TM --- (this is equation 2)

Now we will compare equation 1 and equation 2

TB/TX = TX/TC

TX² = TB x TC (proved)

Answer 2

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$\sf(XM = AB, XN = AC)(XM=AB,XN=AC) [Parallel]$

So,

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf= (BN || XM) = parallel$

USING PROPORTIONALITY THEOREM

$\sf( \frac{tb}{tx} = \frac{tn}{tm} )....eq - (1)$

$\begin{lgathered}\sf = (XN || AC) \\ \sf= (XN || CM)\end{lgathered}$

$\sf = (XN || CM)$

USING PROPORTIONALITY THEOREM

$\sf( \frac{tx}{tc} = \frac{tn}{tm} ).....eq - (2)$

Comparing both equation[ eq - (1) and eq - (2)]

$\sf (\frac{tb}{tx} = \frac{tx}{tb})$

$\sf(TX {}^{2} = TB \: × \: TC)$

$\sf{Therefore ,PROVED}$