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Now, \[\Delta DBE\] and \[\Delta ECD\] being on the same base DE and between the same parallels DE and BC, we have,\[ar(\Delta DBE)=ar(\Delta ECD)\] then we say that the basic proportionality theorem is proved.

Basic proportionality theorem:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Given:

A \[\Delta ABC\]in which \[DE\parallel BC\]and DE intersects AB and AC at D and E respectively.

To prove that:

\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\]

Construction:

Join BE and CD.

Draw \[EL\bot AB\]and \[DM\bot AC\]

Proof:

We have the

\[ar(\Delta ADE)=\dfrac{1}{2}\times AD\times EL\]

\[ar(\Delta DBE)=\dfrac{1}{2}\times DB\times EL\]

Therefore the ratio of these two is \[\dfrac{ar(\Delta ADE)}{ar(\Delta DBE)}=\dfrac{AD}{DB}\]. . . . . . . . . . . . . . (1)

Similarly,

\[ar(\Delta ADE)=ar(\Delta ADE)=\dfrac{1}{2}\times AE\times DM\]

\[ar(\Delta ECD)=\dfrac{1}{2}\times EC\times DM\]

Therefore the ratio of these two is \[\dfrac{ar(\Delta ADE)}{ar(\Delta ECD)}=\dfrac{AE}{EC}\]. . . . . . . . . . . .. . . (2)

Now, \[\Delta DBE\] and \[\Delta ECD\] being on the same base DE and between the same parallels DE and BC, we have,

\[ar(\Delta DBE)=ar(\Delta ECD)\]. . . . . . . . . . . (3)

From equations 1, 2, 3 we can conclude that

\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\]

Hence we can say that the basic proportionality theorem is proved.