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) =1/4 ar (ABC)

(ii) ar (BDE) = ½ ar (BAE)

(iii) ar (ABC) = 2 ar (BEC)

(iv) ar (BFE) = ar (AFD)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) = 1/8 ar (AFC)

Solution:

(i) Assume that G and H are the mid-points of the sides AB and AC respectively.

Join the mid-points with line-segment GH. Here, GH is parallel to third side.

, BC will be half of the length of BC by mid-point theorem.

∴ GH =1/2 BC and GH || BD

∴ GH = BD = DC and GH || BD (Since, D is the mid-point of BC)

Similarly,

GD = HC = HA

HD = AG = BG

, ΔABC is divided into 4 equal equilateral triangles ΔBGD, ΔAGH, ΔDHC and ΔGHD

We can say that,

ΔBGD = ¼ ΔABC

Considering, ΔBDG and ΔBDE

BD = BD (Common base)

Since both triangles are equilateral triangle, we can say that,

BG = BE

DG = DE

, ΔBDG ≅ΔBDE [By SSS congruency]

, area (ΔBDG) = area (ΔBDE)

ar (ΔBDE) = ¼ ar (ΔABC)

Hence proved

(ii)

ar(ΔBDE) = ar(ΔAED) (Common base DE and DE||AB)

ar(ΔBDE)−ar(ΔFED) = ar(ΔAED)−ar (ΔFED)

ar(ΔBEF) = ar(ΔAFD) …(i)

Now,

ar(ΔABD) = ar(ΔABF)+ar(ΔAFD)

ar(ΔABD) = ar(ΔABF)+ar(ΔBEF) [From equation (i)]

ar(ΔABD) = ar(ΔABE) …(ii)

AD is the median of ΔABC.

ar(ΔABD) = ½ ar (ΔABC)

= (4/2) ar (ΔBDE)

= 2 ar (ΔBDE)…(iii)

From (ii) and (iii), we obtain

2 ar (ΔBDE) = ar (ΔABE)

ar (BDE) = ½ ar (BAE)

Hence proved

(iii) ar(ΔABE) = ar(ΔBEC) [Common base BE and BE || AC]

ar(ΔABF) + ar(ΔBEF) = ar(ΔBEC)

From eqn (i), we get,

ar(ΔABF) + ar(ΔAFD) = ar(ΔBEC)

ar(ΔABD) = ar(ΔBEC)

½ ar(ΔABC) = ar(ΔBEC)

ar(ΔABC) = 2 ar(ΔBEC)

Hence proved

(iv) ΔBDE and ΔAED lie on the same base (DE) and are in-between the parallel lines DE and AB.

∴ar (ΔBDE) = ar (ΔAED)

Subtracting ar(ΔFED) from L.H.S and R.H.S,

We get,

∴ar (ΔBDE)−ar (ΔFED) = ar (ΔAED)−ar (ΔFED)

∴ar (ΔBFE) = ar(ΔAFD)

Hence proved

(v) Assume that h is the height of vertex E, corresponding to the side BD in ΔBDE.

Also assume that H is the height of vertex A, corresponding to the side BC in ΔABC.

While solving Question (i),

We saw that,

ar (ΔBDE) = ¼ ar (ΔABC)

While solving Question (iv),

We saw that,

ar (ΔBFE) = ar (ΔAFD).

∴ar (ΔBFE) = ar (ΔAFD)

= 2 ar (ΔFED)

Hence, ar (ΔBFE) = 2 ar (ΔFED)

Hence proved

(vi) ar (ΔAFC) = ar (ΔAFD) + ar(ΔADC)

= 2 ar (ΔFED) + (1/2) ar(ΔABC) [using (v)

= 2 ar (ΔFED) + ½ [4ar(ΔBDE)] [Using result of Question (i)]

= 2 ar (ΔFED) +2 ar(ΔBDE)

Since, ΔBDE and ΔAED are on the same base and between same parallels

= 2 ar (ΔFED) +2 ar (ΔAED)

= 2 ar (ΔFED) +2 [ar (ΔAFD) +ar (ΔFED)]

= 2 ar (ΔFED) +2 ar (ΔAFD) +2 ar (ΔFED) [From question (viii)]

= 4 ar (ΔFED) +4 ar (ΔFED)

⇒ar (ΔAFC) = 8 ar (ΔFED)

⇒ar (ΔFED) = (1/8) ar (ΔAFC)

Hence proved