Theorem 6.2

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Theorem 6.2 : If a line divides any two sides of a triangle in a same ratio then and it is parallel to the third side.

Sollution : -

Given =》In triangle ABC, AB/BD=AE/CE

To prove =》DE || BC

Construction =》Draw DE' || BC

Proof,

Since DE' || BC  (by const.)

by theorem 6.1,

AD/BD = AE'/CE'  ...(1)

But,

given that AD/BD = AE/CE ...(2)

From (1) & (2) we have,

AE'/CE' = AE/CE   ...(3)

Now,

AE' + CE' = AC  ...(a)

AE + EC = AC   ...(b)

Then, add 1 to eq(3) we get

--> AE'/CE' + 1 = AE/CE + 1

-->AE'+CE'/CE' = AE+CE/CE

From eq(a) & (b) we have,

AC/CE' = AC/CE

1/CE'    = 1/CE

or,   CE' = CE

From above we can say that point A and E' are  consideing

therefore,  DE' || BC

i.e.,              DE || BC

Hence, proved