In a $Δ\ ABC,$ $D$ and $E$ are points on the sides $AB$ and $AC$ respectively. For each of the following cases show that $DE\ ∥\ BC$:
$AB\ =\ 10.8\ cm$, $BD\ =\ 4.5\ cm$, $AC\ =\ 4.8\ cm$, and $AE\ =\ 2.8\ cm$.

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Given:

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

$AB\ =\ 10.8\ cm$, $BD\ =\ 4.5\ cm$, $AC\ =\ 4.8\ cm$, and $AE\ =\ 2.8\ cm$.

To do:

We have to prove that $DE\ ∥\ BC$.

Solution:

From the figure,

$AD = AB - BD = 10.8 - 4.5 = 6.3 cm$

$CE = AC - AE = 4.8 – 2.8 = 2 cm$

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,

$\frac{AD}{DB}=\frac{6.3}{4.5}=\frac{7}{5}$

$\frac{AE}{EC}=\frac{2.8}{2}=\frac{7}{5}$

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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