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Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

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Last updated date: 21st Jul 2024
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Answer
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Hint: In the given question, we will draw the triangle and calculate the areas of the triangle using different heights. Then we will take the common ratio of the sides of the triangle to arrive at the answer. We can find the area of the triangle using the formula \[Area\vartriangle = \dfrac{1}{2} \times Base \times Height\] .

Complete step-by-step answer:
Let us draw \[\vartriangle ABC\] with \[\overline {DE} \] parallel to \[\overline {BC} \] intersecting the two sides \[\overline {AB} \] and \[\overline {AC} \] at points \[D\] and \[E\] respectively. Now in order to prove that \[\overline {DE} \] divides the two sides in same ratio i.e. \[\dfrac{{AD}}{{DB}} = \dfrac{{AE}}{{EC}}\] , let us two perpendicular lines - \[\overline {DM} \] perpendicular to \[\overline {AC} \] and \[\overline {EN} \] perpendicular to \[\overline {AB} \] . Diagram is shown below for understanding:
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Consider the triangles \[\vartriangle ADE\] , \[\vartriangle BDE\] and \[\vartriangle DEC\] . We can obtain their area with the help of formula:
 \[Area\vartriangle = \dfrac{1}{2} \times Base \times Height\]
Area of \[\vartriangle ADE = \dfrac{1}{2} \times AD \times EN = \dfrac{1}{2} \times AE \times DM\]
(Since we can take two base and calculate area with two different sides \[\overline {AB} \] and \[\overline {AC} \] )
Area of \[\vartriangle BDE = \dfrac{1}{2} \times DB \times EN\]
Area of \[\vartriangle DEC = \dfrac{1}{2} \times EC \times DM\]
Now to obtain the ratios, we will divide the area of triangles as follows:
 \[\dfrac{{Area\,of\,\vartriangle ADE}}{{Area\,of\,\vartriangle BDE}} = \dfrac{{\dfrac{1}{2} \times AD \times EN}}{{\dfrac{1}{2} \times DB \times EN}}\]
Cancelling the equal terms, we get,
 \[\dfrac{{Area\,of\,\vartriangle ADE}}{{Area\,of\,\vartriangle BDE}} = \dfrac{{AD}}{{DB}}\] --------(1)
Similarly, for \[\vartriangle ADE\] and \[\vartriangle DEC\] the ratio will be as follows:
 \[\dfrac{{Area\,of\,\vartriangle ADE}}{{Area\,of\,\vartriangle DEC}} = \dfrac{{\dfrac{1}{2} \times AE \times DM}}{{\dfrac{1}{2} \times EC \times DM}}\]
 \[\dfrac{{Area\,of\,\vartriangle ADE}}{{Area\,of\,\vartriangle DEC}} = \dfrac{{AE}}{{EC}}\] ---------(2)
We know that \[\vartriangle BDE\] and \[\vartriangle DEC\] have a common base \[\overline {DE} \] which is parallel to \[\overline {BC} \] . The area of triangles having common base and lying between parallel lines will be equal. Hence, we can say that-
 \[Area\,of\,\vartriangle BDE = Area\,of\,\vartriangle DEC\] --------(3)
Therefore, we can conclude from the (1), (2) and (3) that-
 \[\dfrac{{AD}}{{DB}} = \dfrac{{AE}}{{EC}}\] .
Thus, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Note: Alternatively, we can prove as follows using the similarity theorem:
In \[\vartriangle ADE\] and \[\vartriangle ABC\] , \[\angle DAE = \angle BAC\] because \[\angle A\] is common.
 \[\angle ADE = \angle ABC\] since corresponding angles are equal as \[\overline {DE} \] is parallel to \[\overline {BC} \]
Similarly, \[\angle AED = \angle ACB\] .
So, we can conclude that both triangles are similar.
 \[\vartriangle ADE \sim \vartriangle ABC\]
Now in similar triangles, corresponding sides are in similar ratios-
 \[\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}\]
Reversing the ratio, we get,
 \[\dfrac{{AB}}{{AD}} = \dfrac{{AC}}{{AE}}\]
Deducting the denominator from both the sides, we get,
 \[\dfrac{{AB - AD}}{{AD}} = \dfrac{{AC - AE}}{{AE}}\]
 \[\dfrac{{DB}}{{AD}} = \dfrac{{EC}}{{AE}}\]
Reversing the ratio again, we get,
 \[\dfrac{{AD}}{{DB}} = \dfrac{{AE}}{{EC}}\]
Hence proved.