In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

Solution:

(i) Let G and H be the mid-points of side AB and AC respectively.

Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).



$\Rightarrow \mathrm{GH}=\frac{1}{2} \mathrm{BC}$ and $\mathrm{GH} \| \mathrm{BD}$

$\Rightarrow \mathrm{GH}=\mathrm{BD}=\mathrm{DC}$ and $\mathrm{GH} \| \mathrm{BD}(\mathrm{D}$ is the mid-point of $\mathrm{BC})$

Consider quadrilateral GHDB.

$\mathrm{GH} \| \mathrm{BD}$ and $\mathrm{GH}=\mathrm{BD}$

Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.

Therefore, BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram.

We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.

Hence, Area $(\Delta B D G)=$ Area $(\Delta H G D)$

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.

ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)

ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)

ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)

$\operatorname{ar}(\Delta \mathrm{ABC})=\operatorname{ar}(\Delta \mathrm{BDG})+\operatorname{ar}(\Delta \mathrm{GDH})+\operatorname{ar}(\Delta \mathrm{DCH})+\operatorname{ar}(\Delta \mathrm{AGH})$

$\operatorname{ar}(\triangle \mathrm{ABC})=4 \times \operatorname{ar}(\triangle \mathrm{BDE})$

Hence, $\operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC})$

(ii)Area $(\Delta B D E)=$ Area $(\triangle A E D)($ Common base $D E$ and $D E \| A B)$

Area $(\triangle B D E)$ - Area $(\triangle F E D)=$ Area $(\triangle A E D)-$ Area $(\triangle F E D)$

Area $(\triangle B E F)=$ Area $(\triangle A F D)(1)$

Area $(\triangle \mathrm{ABD})=$ Area $(\triangle \mathrm{ABF})+$ Area $(\triangle \mathrm{AFD})$

Area $(\triangle \mathrm{ABD})=$ Area $(\triangle \mathrm{ABF})+$ Area $(\triangle \mathrm{BEF})[$ From equation $(1)]$

Area $(\Delta \mathrm{ABD})=$ Area $(\Delta \mathrm{AB} \mathrm{E})(2)$

$A D$ is the median in $\triangle A B C$.

$\operatorname{ar}(\Delta \mathrm{ABD})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABC})$

$=\frac{4}{2} \operatorname{ar}(\Delta B D E)$

(Asproved earlier)

$\operatorname{ar}(\Delta \mathrm{ABD})=2 \operatorname{ar}(\triangle \mathrm{BDE})$(3)

From $(2)$ and $(3)$, we obtain

$2 \operatorname{ar}(\triangle B D E)=\operatorname{ar}(\triangle A B E)$

Or, ar $(\Delta \mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABE})$



$\operatorname{ar}(\triangle \mathrm{ABE})=\operatorname{ar}(\triangle \mathrm{BEC})($ Common base $\mathrm{BE}$ and $\mathrm{BE} \| \mathrm{AC})$

$\operatorname{ar}(\Delta \mathrm{ABF})+\operatorname{ar}(\Delta \mathrm{BEF})=\operatorname{ar}(\Delta \mathrm{BEC})$

Using equation (1), we obtain

$\operatorname{ar}(\triangle \mathrm{ABF})+\operatorname{ar}(\triangle \mathrm{AFD})=\operatorname{ar}(\triangle \mathrm{BEC})$

$\operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{BEC})$

$\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABC})=\operatorname{ar}(\Delta \mathrm{BEC})$

$\operatorname{ar}(\triangle \mathrm{ABC})=2 \operatorname{ar}(\triangle \mathrm{BEC})$

(iv)It is seen that $\triangle B D E$ and ar $\triangle A E D$ lie on the same base (DE) and between the parallels DE and $A B$.

$\therefore \operatorname{ar}(\triangle \mathrm{BDE})=\operatorname{ar}(\triangle \mathrm{AED})$

$\Rightarrow \operatorname{ar}(\triangle B D E)-\operatorname{ar}(\triangle F E D)=\operatorname{ar}(\triangle A E D)-\operatorname{ar}(\triangle F E D)$

$\therefore \operatorname{ar}(\Delta \mathrm{BFE})=\operatorname{ar}(\triangle \mathrm{AFD})$

(v) Let $h$ be the height of vertex $E$, corresponding to the side BD in $\triangle B D E$.

Let $H$ be the heiaht of vertex $A$ correspondina to the side $B C \operatorname{in} \triangle A B C$

In (i), it was shown that $\operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC})$.

$\therefore \frac{1}{2} \times \mathrm{BD} \times h=\frac{1}{4}\left(\frac{1}{2} \times \mathrm{BC} \times H\right)$

$\Rightarrow \mathrm{BD} \times h=\frac{1}{4}(2 \mathrm{BD} \times H)$

$\Rightarrow h=\frac{1}{2} H$

In (iv) it was shown that $\operatorname{ar}(\Delta B F E)=\operatorname{ar}(\Delta A F D)$.

$=\operatorname{ar}(\mathrm{BFE})+\frac{1}{2} \operatorname{ar}(\mathrm{ABC})$

$[$ In (vi), $\operatorname{ar}(\mathrm{BFE})=\operatorname{ar}(\mathrm{AFD}) ; \mathrm{AD}$ is median of $\triangle \mathrm{ABC}]$

$=\operatorname{ar}(\mathrm{BFE})+\frac{1}{2} \times 4 \operatorname{ar}(\mathrm{BDE})$

$\left[\operatorname{In}(i), \operatorname{ar}(B D E)=\frac{1}{4} \operatorname{ar}(A B C)\right]$

$=\operatorname{ar}(\mathrm{BFE})+2 \operatorname{ar}(\mathrm{BDE})$..(5)

Now, by $(v), \operatorname{ar}(B F E)=2 \operatorname{ar}(F E D) \ldots(6)$

$\operatorname{ar}(\mathrm{BDE})=\operatorname{ar}(\mathrm{BFE})+\operatorname{ar}(\mathrm{FED})=2 \operatorname{ar}(\mathrm{FED})+\operatorname{ar}(\mathrm{FED})=3 \operatorname{ar}(\mathrm{FED})$..(7)

Therefore, from equations $(5),(6)$, and $(7)$, we get:

$\operatorname{ar}(\mathrm{AFC})=2 \operatorname{ar}(\mathrm{FED})+2 \times 3 \operatorname{ar}(\mathrm{FED})=8 \operatorname{ar}(\mathrm{FED})$

$\therefore \operatorname{ar}(\mathrm{AFC})=8 \operatorname{ar}(\mathrm{FED})$

Hence, $\operatorname{ar}(\mathrm{FED})=\frac{1}{8} \operatorname{ar}(\mathrm{AFC})$