Question

X is any point inside $△$ABC. XA, XB and XC are joined. 'E' is any point on AX. If EF || AB, FG || BC. Prove that EG || AC.

Answer 1

The converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

In $△XAB$, $EF||AB$, therefore,

$\frac{XE}{EA}=\frac{XF}{FB}.......\left(1\right)$

In $△XBC$, $FG||BC$, therefore,

$\frac{XF}{FB}=\frac{XG}{GC}.......\left(2\right)$

Comparing equations 1 and 2, we get

$\frac{XE}{EA}=\frac{XG}{GC}$

Now in $△XCA$, if we use the converse of the basic proportionality theorem then we get that $EG||AC$.

Hence, $EG||AC$.