Question

X is a point on the side BC of triangleABC .XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MAN produced meets CB produced at T. Prove that TX2=TB.TC

Answer 1

Step-by-step explanation:

From figure:

(i)

From ΔTXM, we have

⇒ BN ║ XM.

By using basic proportionality theorem, we have

⇒ (TB/TX) = (TN/TM)

(ii)

From ΔTMC, we have

⇒ XN ║ CM.

By using basic proportionality theorem, we have

⇒ (TX/TC) = (TN/TM).

On comparing (i) & (ii), we get

⇒ (TB/TX) = (TX/TC)

⇒ TX² = TB * TC.

Hope it helps!

Answer 2

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)