Question

XY is a line parallel to the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$. If $\mathrm{BE}\parallel \mathrm{AC}$ and $\mathrm{CF}\parallel \mathrm{AB}$ meet $\mathrm{XY}$ at $\mathrm{E}$ and $\mathrm{F}$ respectively, show that $\mathrm{ar}(∆\mathrm{ABE})=\mathrm{ar}(∆\mathrm{ACF})$.

Answer 1

Step 1:Plot a parallelogram diagram:

When a triangle and a parallelogram share the same base and are connected by the same parallel lines, the triangle's area is half that of the parallelogram.

Furthermore, if two parallelograms are placed on the same base and between the same pair of parallel lines, they will have the same area.

Let's draw points $\mathrm{X}$ and $\mathrm{Y}$ on sides, $\mathrm{AB}$ and $\mathrm{AC}$, respectively, crossed by a line $\mathrm{EF}$.

Step 2:Determine the parallelogram

Let's consider $\mathrm{BCYE}$
It is given that, $\mathrm{XY}\parallel \mathrm{BC}$ so, $\mathrm{EY}\parallel \mathrm{BC}$
Also,$\mathrm{BE}\parallel \mathrm{AC}$ so, $\mathrm{BE}\parallel \mathrm{CY}$
Therefore, $\mathrm{BCYE}$ is a parallelogram.
Similarly, In $\mathrm{BCFX}$
It is given that, $\mathrm{XY}\parallel \mathrm{BC}$ so, $\mathrm{XF}\parallel \mathrm{BC}$
Since, $\mathrm{CF}\parallel \mathrm{AB}$ so, $\mathrm{CF}\parallel \mathrm{BX}$
Therefore, $\mathrm{BCFX}$ is a parallelogram.
Parallelograms $\mathrm{BCYE}$ and $\mathrm{BCFX}$ are lying on the same base $\mathrm{BC}$ and between the same parallels $\mathrm{BC}$ and $\mathrm{EF}$.

Step 3:Prove the given area's

According to theorem Parallelograms on the same base and between the same parallels are equal in area.

$\therefore \mathrm{Area}\left(\mathrm{BCYE}\right)=\mathrm{Area}\left(\mathrm{BCFX}\right)...\left(1\right)$

Now, Consider parallelogram $\mathrm{BCYE}$ and $\mathrm{\Delta AEB}$

They are lying on the same base $\mathrm{BE}$ and are between the same parallels $\mathrm{BE}$ and $\mathrm{AC}$.

$\therefore \mathrm{Area}\left(\mathrm{\Delta ABE}\right)=\frac{1}{2}\mathrm{Area}\left(\mathrm{BCYE}\right)...\left(2\right)$

Also, parallelogram BCFX and $\mathrm{\Delta ACF}$ are lying on the same base $\mathrm{CF}$ and existing between the same parallels $\mathrm{CF}$ and $\mathrm{AB}$.

$\therefore \mathrm{Area}\left(\mathrm{\Delta ACF}\right)=\frac{1}{2}\mathrm{Area}\left(\mathrm{BCFX}\right)...\left(3\right)$

From Equations $\left(1\right),\left(2\right)$, and $\left(3\right)$, we obtain

$\mathrm{Area}\left(\mathrm{\Delta ABE}\right)=\mathrm{Area}\left(\mathrm{\Delta ACF}\right)$

Hence, it's proved that $\mathbf{Area}\mathbf{}\mathbf{\left(}\mathbf{\Delta ABE}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{Area}\mathbf{}\mathbf{\left(}\mathbf{\Delta ACF}\mathbf{\right)}$