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 TX^2=TB×TC...
Plz tell me...

Answer 1

hey mate
here's the proof

Answer 2

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)