Question -
Answer -
(i) Assume that G andH are the mid-points of the sides AB and AC respectively.
Join the mid-pointswith line-segment GH. Here, GH is parallel to third side.
, BC will be half ofthe 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 into4 equal equilateral triangles ΔBGD, ΔAGH, ΔDHC and ΔGHD
We can say that,
ΔBGD = ¼ ΔABC
Considering, ΔBDG andΔBDE
BD = BD (Common base)
Since both trianglesare equilateral triangle, we can say that,
BG = BE
DG = DE
, ΔBDG ΔBDE [By SSScongruency]
, 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 ofΔABC.
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 lieon 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 isthe height of vertex E, corresponding to the side BD in ΔBDE.
Also assume that H isthe 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) = 2ar (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) +2ar(ΔBDE)
Since, ΔBDE and ΔAEDare 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