X is a point on the side BC of AABC. XM and XN are drawn parallel to A

1.	X is a point on the side BC of AABC. XM and XN are drawn parallel to AB and ACrespectively meeting AB in N and AC in M. MN produced meets CB produced at T.Prove that TX2-TB TC
Answer» 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)

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

Answer»

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)