Maths

124 MATHEMATICS
Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same
ratio.
Proof : We are given a triangle ABC in which a line
parallel to side BC intersects other two sides AB and
AC at D and E respectively (see Fig. 6.10).
We need to prove that AD AE
DB EC .
Let us join BE and CD and then draw DM ✁ AC and
EN ✁ AB.
Fig. 6.10
Now, area of ✂ ADE (= 1 2 base × height) = 1 2
AD × EN.
Recall from Class IX, that area of ✂ ADE is denoted as ar(ADE).
So,
ar(ADE) = 1 2
AD × EN
Similarly,
ar(BDE) = 1 2
DB × EN,
ar(ADE) = 1 2 AE × DM and ar(DEC) = 1 2
EC × DM.
Therefore,
1
ar(ADE) AD × EN AD
ar(BDE) = 2
(1)
✄
1
DB × EN
DB
2
and
1
ar(ADE) AE × DM AE
ar(DEC) = 2
(2)
✄
1
EC × DM
EC
2
Note that ✂ BDE and DEC are on the same base DE and between the same parallels
BC and DE.
So, ar(BDE) = ar(DEC) (3)