###### Pawan Prajapati

Last Activity: 2 Years ago

Hint: To prove this theorem first we will join BE and CD. Then draw a line EL perpendicular to AB and line DM perpendicular to AC. Now we will find the ratio of area of Δ
ADE to Δ
DBE and ratio of area of Δ
ADE to Δ
ECD. Comparing the ratios we will get the final answer.
Complete step-by-step answer:
Now, ΔDBE
and ΔECD
being on the same base DE and between the same parallels DE and BC, we have,ar(ΔDBE)=ar(ΔECD)
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 ΔABC
in which DE∥BC
and DE intersects AB and AC at D and E respectively.
To prove that:
ADDB=AEEC
Construction:
Join BE and CD.
Draw EL⊥AB
and DM⊥AC
Proof:
We have the
ar(ΔADE)=12×AD×EL
ar(ΔDBE)=12×DB×EL
Therefore the ratio of these two is ar(ΔADE)ar(ΔDBE)=ADDB
. . . . . . . . . . . . . . (1)
Similarly,
ar(ΔADE)=ar(ΔADE)=12×AE×DM
ar(ΔECD)=12×EC×DM
Therefore the ratio of these two is ar(ΔADE)ar(ΔECD)=AEEC
. . . . . . . . . . . .. . . (2)
Now, ΔDBE
and ΔECD
being on the same base DE and between the same parallels DE and BC, we have,
ar(ΔDBE)=ar(ΔECD)
. . . . . . . . . . . (3)
From equations 1, 2, 3 we can conclude that
ADDB=AEEC
Hence we can say that the basic proportionality theorem is proved.
Note: The formula for area of the triangle is given by 12×b×h
where b, h are base and height respectively. If two triangles are on the same base and between the same parallels then the area of those two triangles are equal.