In a $\vartriangle ABC$, D and E are points on the sides AB and AC respectively. For each of the following cases show that $DE||BC$: iii) $AB=10.8cm$, $BD=4.5cm$, $AC=4.8cm$, and $AE=2.8cm$.iv) $AD=5.7cm$, $BD=9.5cm$, $AE=3.3cm$, and $EC=5.5cm$.

(iii)

Given information: $=10.8 cm,$$BD=4.5cm$, $AC=4.8cm$, and $AE=2.8cm$.

Required to prove: $DE||BC$.

Proof:

$AD=AB-DB=10.8-4.5=6.3$

And,

$CE=AC-AE=4.8-2.8=2$

As ,we can see that

$AD/BD=6.3/4.5=2.8/2.0=AE/CE=7/5$

Hence, by using the converse of Thale’s Theorem

So, we can say that,

$DE||BC$

Hence Proved

(iv)

Given information= $AD=5.7cm$, $BD=9.5cm$, $AE=3.3cm$, and $EC=5.5cm$

Required to prove: $DE||BC$

Proof:

$AD/BD=5.7/9.5=3/5$

And,

$AE/CE=3.3/5.5=3/5$

Thus,

$AD/BD=AE/CE$z

Hence, by using the converse of Thale’s Theorem

We have,

$DE||BC$

Hence Proved.