NCERT Solutions
Class 9 Maths
Areas of Parallelograms and Triangles
Ex.9.4 Q.5
In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:
(i) ar (BDE) = 
(ii) ar (BDE) = 
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 
[ Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.
(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).
GH = 
and GH || BD
GH = BD = DC and GH || BD
[D is the mid-point of BC]
Consider quadrilateral GHDB,
GH ||BD and GH = 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 (∆BDG) = Area (∆HGD)
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(∆GDH) = ar(∆CHD)
[For parallelogram DCHG]
ar(∆GDH) = ar(∆HAG)
[For parallelogram GDHA]
ar(∆BDE) = ar(∆DBG)
[For parallelogram BEDG]
ar(∆ABC) = ar(∆BDG) + ar(∆GDH) + ar(∆DCH) + ar(∆AGH)
ar(∆ABC) = 4 × ar(∆BDE)
Hence, ar(∆BDE) = 
(ii)Area (∆BDE) = Area (∆AED)
[Common base DE and DE||AB]
Area (∆BDE) − Area (∆FED) = Area (∆AED) − Area (∆FED)
Area (∆BEF) = Area (∆AFD) …………. (1)
Area (∆ABD) = Area (∆ABF) + Area (∆AFD)
Area (∆ABD) = Area (∆ABF) + Area (∆BEF)
[From equation (1)]
Area (∆ABD) = Area (∆ABE) ………… (2)
AD is the median in ∆ABC.
ar(∆ABD) = 
= 4 × 
[As proved earlier]
= 2 × ar(∆BDE)
=> ar(∆ABD) = 2 × ar(∆BDE) ………... (3)
From equation (2) and (3), we get
2 × ar(∆BDE) = ar(∆ABE)
=> ar(∆BDE) = 
(iii) ar(∆ABE) = ar(∆BEC)
[Common base BE and BE||AC]
ar(∆ABF) + ar(∆BEF) = ar(∆BEC)
Using equation (1), we obtain
ar(∆ABF) + ar(∆AFD) = ar(∆BEC)
=> ar(∆ABD) = ar(∆BEC)
=> = ar(∆BEC)
=> ar(∆ABC) = 2 × ar(∆BEC)
(iv) It is seen that ∆BDE and ar ∆AED lie on the same base (DE)
and between the parallels DE and AB.
ar (∆BDE) = ar (∆AED)
ar (∆BDE) − ar (∆FED) = ar (∆AED) − ar (∆FED)
ar (∆BFE) = ar (∆AFD)
(v) Let h be the height of vertex E, corresponding to the side BD in ∆BDE.
Let H be the height of vertex A, corresponding to the side BC in ∆ABC.
In (i), it was shown that ar(∆BDE) =
=> 
× BD × h = ( × BC × H) ÷ 4
=> BD × h = (2BD × H) ÷ 4
=> h = 
In (iv), it was shown that ar (∆BFE) = ar (∆AFD).
ar (∆BFE) = ar (∆AFD)
= × FD × H
= × FD × 2H
= 2(
× FD × H)
= 2 × ar (∆FED)
Hence, ar (∆BFE) = 2 × ar (∆FED)
(vi) Area (AFC) = area (AFD) + area (ADC)
[in (iv), ar(∆BFE) = ar(∆AFD), AD is the median of ΔABC]
= ar (∆BFE) +
= ar (∆BFE) + 4 × 
= ar (∆BFE) + 2 × ar (∆BDE) …………... (5)
Now, by (v), ar (∆BFE) = 2 × ar (∆FED) …………... (6)
ar (∆BDE) = ar (∆BFE) + ar (∆FED) = 2 × ar (∆FED) + ar (∆FED)
= 3 × ar (∆FED) ……… (7)
From equation (5), (6) and (7), we get
ar (∆AFC) = 2 × ar (∆FED) + 2 × 3 × ar (∆FED)
= 2 × ar (∆FED) + 6 × ar (∆FED)
= 8 × ar (∆FED)
Hence, ar (∆FED) = 