Question

State and prove basic proportionality theorem.

Answer 1

Hint: The basic proportionality theorem states that “If a line is parallel to one side of a triangle which intersects the other into two distinct points, then the line divides those sides of a triangle in proportion”.

Complete answer:
For a triangle shown below

Here, if \[DE\parallel BC\] then, \[\dfrac{AD}{DB}=\dfrac{AE}{EC}\]
We prove this theorem by taking the suitable constructions and using some other standard theorems.
We are asked to state and prove the basic proportionality theorem.
We know that the basic proportionality theorem states that “If a line is parallel to one side of a triangle which intersects the other into two distinct points, then the line divides those sides of a triangle in proportion”.
Now, let us take a triangle \[\Delta ABC\] and \[DE\parallel BC\]
Now, let us join the points B to E and C to D, also let us draw the heights of the triangle \[\Delta ADE\] as shown below

Here, we can see that the line segment EF is height to both triangles \[\Delta ADE\] and \[\Delta BDE\]
We know that the formula of area of the triangle as
\[A=\dfrac{1}{2}\left( base \right)\left( height \right)\]
By using the above formula let us find the ratio of area of triangle \[\Delta ADE\] and \[\Delta BDE\] then we get
\[\begin{align}
  & \Rightarrow \dfrac{ar\left( \Delta ADE \right)}{ar\left( \Delta BDE \right)}=\dfrac{\dfrac{1}{2}\times AD\times EF}{\dfrac{1}{2}\times DB\times EF} \\
 & \Rightarrow \dfrac{ar\left( \Delta ADE \right)}{ar\left( \Delta BDE \right)}=\dfrac{AD}{DB}......equation(i) \\
\end{align}\]
Similarly, we can see that the line segment DG is the height of both triangles \[\Delta ADE\] and \[\Delta CDE\]
By using the area formula of triangle let us find the ratio of area of triangle \[\Delta ADE\] and \[\Delta CDE\] then we get
\[\begin{align}
  & \Rightarrow \dfrac{ar\left( \Delta ADE \right)}{ar\left( \Delta CDE \right)}=\dfrac{\dfrac{1}{2}\times AE\times DG}{\dfrac{1}{2}\times CE\times DG} \\
 & \Rightarrow \dfrac{ar\left( \Delta ADE \right)}{ar\left( \Delta CDE \right)}=\dfrac{AE}{EC}......equation(ii) \\
\end{align}\]
Here, we can see that the triangles \[\Delta BDE\] and \[\Delta CDE\] have the same base and lie between the same parallel lines.
We know that the condition that two triangles having the same base and lie between same parallel lines have equal area.
By using the above condition we can say that the area of \[\Delta BDE\] and \[\Delta CDE\] are equal that is
\[\Rightarrow ar\left( \Delta BDE \right)=ar\left( \Delta CDE \right)\]
Let us divide the both sides with area of triangle \[\Delta ADE\] then we get
\[\begin{align}
  & \Rightarrow \dfrac{ar\left( \Delta BDE \right)}{ar\left( \Delta ADE \right)}=\dfrac{ar\left( \Delta CDE \right)}{ar\left( \Delta ADE \right)} \\
 & \Rightarrow \dfrac{ar\left( \Delta ADE \right)}{ar\left( \Delta BDE \right)}=\dfrac{ar\left( \Delta ADE \right)}{ar\left( \Delta CDE \right)} \\
\end{align}\]
Now, by substituting the required values from equation (i) and equation (ii) in above equation we get
\[\Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{EC}\]
Therefore, we can conclude that the required result has been proved.
That is for a triangle \[\Delta ABC\], if \[DE\parallel BC\] then, \[\dfrac{AD}{DB}=\dfrac{AE}{EC}\]


Note:
We can prove the above result in other method

Here, let us consider the triangles \[\Delta ADE\] and \[\Delta ABC\]
Here, we can see that all the angles in both triangles \[\Delta ADE\] and \[\Delta ABC\] are equal that is
(1) \[\angle DAE=\angle BAE\left( \because \text{common angle} \right)\]
(2) \[\angle ADE=\angle ABC\left( \because DE\parallel BC \right)\]
(3) \[\angle AED=\angle ACB\left( \because DE\parallel BC \right)\]
We know that if three angles of two triangles are equal then the two triangles are similar that is
\[\Rightarrow \Delta ADE\sim \Delta ABC\]
We know that the condition of similar triangles that is the corresponding sides are in proportion.
By using the above condition to triangles \[\Delta ADE,\Delta ABC\] then we get
\[\Rightarrow \dfrac{AB}{AD}=\dfrac{AC}{AE}\]
Here, we can see that the sides AB and AC are divided in to two parts.
By rewriting the sides AB and AC as sum of divides parts then we get
\[\begin{align}
  & \Rightarrow \dfrac{AD+DB}{AD}=\dfrac{AE+EC}{AE} \\
 & \Rightarrow 1+\dfrac{DB}{AD}=1+\dfrac{EC}{AE} \\
 & \Rightarrow \dfrac{DB}{AD}=\dfrac{EC}{AE} \\
 & \Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{EC} \\
\end{align}\]
Therefore, we can conclude that the required result has been proved.
That is for a triangle \[\Delta ABC\], if \[DE\parallel BC\] then, \[\dfrac{AD}{DB}=\dfrac{AE}{EC}\]