Question

In a triangle $ABC$, $E$ is a midpoint of median $AD$, show that area of $\Delta ABE$= area of $\Delta ACE$.

Answer 1

Hint: We know that the median of any triangle divides it into two equal parts. It means $AD$ divides $\Delta ABC$ into two equal triangles and $BE$ divides $\Delta ABD$ into two equal parts. Also, $CE$ divides $\Delta ACD$ into two equal parts. From this, we get $\Delta ABE=\Delta ACE$.

Complete step-by-step answer:

It is given in the question that $ABC$ is a triangle in which $E$ is the midpoint of median $AD$ then we have to show that $\Delta ABE=\Delta ACE$.

As we know the median divides any triangle into two equal parts we will use the basic property of the median to show that, area of$(\Delta ABE)=(\Delta ACE)$. As the median divides the triangle into two equal parts it means $AD$ divides $\Delta ABC$ into two equal triangles.

Area of $\Delta ABD=\Delta ACD$ or we can write as $\Delta ABD=\dfrac{1}{2}(\Delta ABC)$……….(i)

Similarly, it is given in the question that $E$ is a midpoint of median $AD$ it means $BE$ is a median of $\Delta ABD$. So, we get the median $BE$ which divides the area of $\Delta ABD$ into two equal parts.

Therefore, area of $\Delta ABE=\Delta BED$ or area of $\Delta ABE=\dfrac{1}{2}(\Delta ABD)$……….(ii)

We know that $\Delta ABD$ is $\dfrac{1}{2}$ of $\Delta ABC$ then, we get

$\Rightarrow $Area of $\Delta ABE=\dfrac{1}{2}\times \dfrac{1}{2}\times \Delta ABC$

$\Rightarrow $Area of $\Delta ABE=\dfrac{1}{4}\times \Delta ABC$……….(iii)

Similarly, in $\Delta ACD$ we have $CE$ as a median. Which means $CE$ divides $\Delta ACD$ into two equal parts.

Since, the area of $\Delta ACE=\Delta CDE$ or it can be written as the area of $\Delta ACE=\dfrac{1}{2}\times \Delta ADC$. Also, we know that $\Delta ADC$ is half of $\Delta ABC$, we get

$\Rightarrow $Area of $\Delta ACE=\dfrac{1}{2}\times \dfrac{1}{2}\times \Delta ABC$

$\Rightarrow $Area of $\Delta ACE=\dfrac{1}{4}\times \Delta ABC$……….(iv)

From equation (iii) and (iv), we get

Area of $\Delta ABE=\Delta ACE$. Hence proved.


Note: We can also solve this question by showing triangle $ABC$ is congruent to triangle $ACE$. Because the area of congruent triangles are always equal. While solving this question students may skip to multiply $\dfrac{1}{2}$ with area of $\Delta ACD$ and area of $\Delta ABD$. Which gives an incorrect answer.