Answer

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**Hint:**Here, we will be proceeding by making the two triangles i.e., $\vartriangle {\text{ADE}}$ and $\vartriangle {\text{ABC}}$ as congruent triangles with the help of Inverse of Basic Proportionality Theorem and concept of corresponding angles.

**Complete step-by-step answer:**

Given, AD=8 cm, DB=12 cm, AE=6 cm and CE=9 cm

To prove: ${\text{BC}} = \dfrac{{5\left( {{\text{DE}}} \right)}}{2}$

If we observe the ratio of AD to DB, it is equal to

$ \dfrac{AD}{DB} = \dfrac{8}{{12}} $

$ \Rightarrow \dfrac{AD}{{DB}} = \dfrac{2}{3} $

Now observe the ratio of AE to CE, it is equal to

$ \dfrac{{{\text{AE}}}}{{{\text{CE}}}} = \dfrac{6}{9} $

$ \Rightarrow \dfrac{{{\text{AE}}}}{{{\text{CE}}}} = \dfrac{2}{3}{\text{ }} \to {\text{(2)}} $

As we know that the Inverse of Basic Proportionality Theorem states that if a line divides two sides of a triangle at distinct points in the same ratio then that line is parallel to the third side of the triangle.

By equations (1) and (2), we can say the ratio in which line DE is dividing the two sides AB and AC of the triangle ABC is equal. So, according to the Inverse of Basic Proportionality Theorem this line DE should be parallel to the third side (i.e., BC) of the triangle ABC.

So, DE is parallel to BC i.e., ${\text{DE}}\parallel {\text{BC}}$

Also we know that if ${\text{DE}}\parallel {\text{BC}}$ then the corresponding angles will be equal.

i.e., $\angle {\text{ADE}} = \angle {\text{ABC}}$ and $\angle {\text{AED}} = \angle {\text{ACB}}$

In triangles $\vartriangle {\text{ADE}}$ and$\vartriangle {\text{ABC}}$,

$\angle {\text{ADE}} = \angle {\text{ABC}}$

$\angle {\text{AED}} = \angle {\text{ACB}}$

$\angle {\text{A}} = \angle {\text{A}}$ common to both the triangles $\vartriangle {\text{ADE}}$ and $\vartriangle {\text{ABC}}$

Therefore, by AAA (Angle-Angle-Angle) congruence rule, we can say that both of these triangles are congruent to each other.

i.e., $\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}}$

Also we know that, if the two triangles are congruent to each other then, the ratio of their corresponding sides will also be equal.

i.e.,

$ \dfrac{{{\text{AD}}}}{{{\text{AB}}}} = \dfrac{{{\text{DE}}}}{{{\text{BC}}}}$

$ \Rightarrow \dfrac{{{\text{AD}}}}{{{\text{AD + DB}}}} = \dfrac{{{\text{DE}}}}{{{\text{BC}}}} $

$ \Rightarrow \dfrac{{\text{8}}}{{{\text{8 + 12}}}} = \dfrac{{{\text{DE}}}}{{{\text{BC}}}} $

$ \Rightarrow \dfrac{{{\text{DE}}}}{{{\text{BC}}}} = \dfrac{{\text{8}}}{{{\text{20}}}} = \dfrac{{\text{2}}}{{\text{5}}} $

$ \Rightarrow {\text{BC}} = \dfrac{{{\text{5}}\left( {{\text{DE}}} \right)}}{{\text{2}}} $

The above equation represents the required relation between BC and DE.

**Note:**In this particular problem, we have considered the ratios $\dfrac{{{\text{AD}}}}{{{\text{AB}}}}$ and $\dfrac{{{\text{DE}}}}{{{\text{BC}}}}$ to be equal in order to obtain the required relationship between BC and DE but we will be getting same results if we would have considered the ratios $\dfrac{{{\text{AE}}}}{{{\text{AC}}}}$ and $\dfrac{{{\text{DE}}}}{{{\text{BC}}}}$ to be equal because $\dfrac{{{\text{AE}}}}{{{\text{AC}}}} = \dfrac{{{\text{AE}}}}{{{\text{AE}} + {\text{CE}}}} = \dfrac{{\text{6}}}{{{\text{6}} + {\text{9}}}} = \dfrac{6}{{15}} = \dfrac{2}{5} = \dfrac{{{\text{AD}}}}{{{\text{AB}}}}$.

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