Question

Prove that the two triangles having the same base (or equal bases) and equal areas lie between the parallel lines.

Answer 1

Hint: The rough figure that represents the given information is shown below.

Here we are given that \[\Delta ABC\] and \[\Delta ABD\] have equal areas.
We solve this problem by constructing the heights of both the triangles as shown in the figure.
Then we use the condition that the lines perpendicular to the same line are parallel to each other and the areas of two triangles are equal condition to prove that CEFD is a parallelogram.
Complete step by step answer:
We use the formula of area of a triangle that is
\[A=\dfrac{1}{2}\left( base \right)\left( height \right)\]
We have the condition that if two opposite sides of a quadrilateral are parallel to each other and are equal then that quadrilateral is said to be parallelogram and in a parallelogram opposite sides are parallel to each other.
We are given that two triangles have the same base and areas.
Let us assume that the two triangles are \[\Delta ABC\] and \[\Delta ABD\] as shown in the figure.
Here, we can see that heights of two triangles as perpendicular to base AB that is
\[\Rightarrow CE\bot AB,DF\bot AB\]
We know that the condition that if two opposite sides of a quadrilateral are parallel to each other
By using the above condition we get
\[\Rightarrow CE\parallel DF........equation(i)\]
Now, let us consider the triangles \[\Delta ABC\] and \[\Delta ABD\]
We know that the formula of area of a triangle that is
\[A=\dfrac{1}{2}\left( base \right)\left( height \right)\]
By using the above formula to triangle \[\Delta ABC\] we get
\[\Rightarrow ar\left( \Delta ABC \right)=\dfrac{1}{2}\times AB\times CE\]
Similarly, by using the area formula to triangle \[\Delta ABD\] we get
\[\Rightarrow ar\left( \Delta ABD \right)=\dfrac{1}{2}\times AB\times DF\]
We are given that the areas of triangles are equal that is
\[\Rightarrow ar\left( \Delta ABC \right)=ar\left( \Delta ABD \right)\]
By substituting the value of areas in above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}\times AB\times CE=\dfrac{1}{2}\times AB\times DF \\
 & \Rightarrow CE=DF.....equation(ii) \\
\end{align}\]
Now, let us consider the quadrilateral CEFD
We know that the condition that if two opposite sides of a quadrilateral are parallel to each other and are equal then that quadrilateral is said to be parallelogram
Now, from equation (i) and equation (ii) we can say that quadrilateral CEFD is a parallelogram.
We also know that in a parallelogram opposite sides are parallel to each other.
By using the above statement we can say that
\[\Rightarrow CD\parallel AB\]
Therefore we can conclude that the triangles \[\Delta ABC\] and \[\Delta ABD\] lie between the parallel lines CD and AB.
Hence the required result has been proved.
Note:
The figure that represents the given information can also be taken as

But here we have one other condition that is
\[\Rightarrow AB=BG\]
This is because in the given statement we are also mentioned that equal base.
The equal base condition will be in the form as shown above.
Here, also we get the same situation that
\[\Rightarrow CE\parallel DF........equation(i)\]
But we may have some changes in the area as follows
\[\Rightarrow ar\left( \Delta ABC \right)=ar\left( \Delta ABD \right)\]
By substituting the value of areas in above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}\times AB\times CE=\dfrac{1}{2}\times BG\times DF \\
 & \Rightarrow CE=DF.....equation(ii) \\
\end{align}\]
Whatever the condition the result is same that is the two triangles having the same base (or equal bases) and equal areas lie between the parallel lines.