In Figure, 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) = \(\frac{1}{4}\) ar (ABC)
(ii) ar (BDE) = \(\frac{1}{2}\) ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = \(\frac{1}{8}\) ar (AFC)
(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 = 1/4 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) = 1/4 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) = 1/2 ar (ΔABC)
= (4/2) ar (ΔBDE)
= 2 ar (ΔBDE) ...…(iii)
From (ii) and (iii), we obtain
2 ar (ΔBDE) = ar (ΔABE)
ar (BDE) = 1/2 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)
1/2 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) = 1/4 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) + 1/2 [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