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, $${\displaystyle \frac{AD}{DB}}={\displaystyle \frac{AE}{EC}}$$
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 $$\mathrm{\Delta }ABC$$ 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 $$\mathrm{\Delta }ADE$$ as shown below

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


Note:
We can prove the above result in other method

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