Question

X is a point on the side bc of triangle abc. Xm and xn are draw parallel to ab and ac respectively meeting ab in n and ac in m. Mn produced meets cb produced at t. Prove that tx square=tb*tc

Answer 1

Answer:

Step-by-step explanation:Hey ur answer is here

Answer 2

Answer:

Step-by-step explanation:

Consider in ΔABC, X is the point on BC. MN meets BC in point T  

Prove that $(XT)^{2}=TB\times{TC}$

Given $XM\parallel{AB}$ and $XN\parallel{AC}$

In ΔTMC and ΔTNX

∠CTM=∠XTN are common angle

$XN\parallel{AC}$

therefore, $XN\parallel{MC}$ and NT intersect XN and MC so

∠XNT=∠CMT

$XN\parallel{AC}$

therefore, $XN\parallel{MC}$ and XT intersect XN and MC so

∠NXT=∠MCT

Therefore

ΔTMC≅ΔTNX

Applying the basic proportionality theorem we get,

$\frac{XT}{TC}=\frac{TN}{TM}$....................equation -1

In ΔTMX and ΔTNB

∠XTM=∠BTN are common angle

$XM\parallel{AB}$

therefore, $XM\parallel{BN}$ and NT intersect BN and XM so

∠XMT=∠BNT

$XM\parallel{AB}$

therefore, $XM\parallel{BN}$ and BT intersect BN and XM so

∠NBT=∠MXT

Therefore

ΔTMC≅ΔTNX

Applying the basic proportionality theorem we get,

$\frac{XT}{TB}=\frac{TM}{TN}$  ....................equation -2

Now multiplying equation-1 and equation-2 we get

$\frac{XT}{TB}\times\frac{XT}{TC}=\frac{TM}{TN}\times\frac{TN}{TM}\\\\\Rightarrow\frac{{XT}^{2}}{TB\times{TC}}=1\\\\\Rightarrow{XT}^{2}=TB\times{TC}$

therefore RHS=LHS (Proved)