Question

X is a point on the side bc of triangle abc .Xm and xz 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×tx=tb×tc

Answer 1

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)