In a $Δ\ ABC,$ $D$ and $E$ are points on the sides $AB$ and $AC$ respectively. For each of the following cases show that $DE\ ∥\ BC$:
$AB\ =\ 5.6\ cm$, $AD\ =\ 1.4\ cm$, $AC\ =\ 7.2\ cm$, and $AE\ =\ 1.8\ cm$.

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Given:

In a $Δ\ ABC,$ $D$ and $E$ are points on the sides $AB$ and $AC$ respectively.

$AB\ =\ 5.6\ cm$, $AD\ =\ 1.4\ cm$, $AE\ =\ 1.8\ cm$, and $AC\ =\ 7.2\ cm$.

To do:

We have to prove that $DE\ ∥\ BC$.

Solution:

From the figure,

$BD = AB - AD = 5.6 - 1.4 = 4.2 cm$

$CE = AC - AE = 7.2 – 1.8 = 5.4 cm$

The Converse of the Basic proportionality theorem states that "If a line divides any of the two sides of a triangle in the same ratio, then that line is parallel to the third side". 

Here,

$\frac{AD}{DB}=\frac{1.4}{4.2}=\frac{1}{3}$

$\frac{AE}{EC}=\frac{1.8}{5.4}=\frac{1}{3}$

Therefore,

$\frac{AD}{DB}=\frac{AE}{EC}$  

Hence, by the converse of Basic proportionality theorem,

$DE\ ∥\ BC$.

Hence proved. 

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Updated on: 10-Oct-2022

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