(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 = 1/2 BC 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 (ΔBDG) = Area (ΔHGD)
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)
ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)
ar (ΔABC) = 4 × ar(ΔBDE)
Hence, ar (BDE) = 1/4 ar (ABC)
(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)
Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)
Area (ΔBEF) = Area (ΔAFD) (1)
Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)
Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]
Area (ΔABD) = Area (ΔABE) (2)
AD is the median in ΔABC.
ar (ΔABD) = 1/2 ar (ΔABC) = 4/2 ar (ΔBDE) (As proved earlier)
ar (ΔABD) = 2 ar (ΔBDE) ......... (3)
From (2) and (3), we obtain
2 ar (ΔBDE) = ar (ΔABE)
Or, ar (ΔBDE) = 1/2 ar (ΔABE)
ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)
ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)
Using equation (1), we obtain
ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)
ar (ΔABD) = ar (ΔBEC)
1/2 ar (ΔABC) = ar (ΔBEC)
ar (ΔABC) = 2 ar (ΔBEC)
(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.
⊥ar (ΔBDE) = ar (ΔAED)
⊥ ar (ΔBDE) − ar (ΔFED) = ar (ΔAED) − ar (ΔFED)
⊥ar (ΔBFE) = ar (ΔAFD)
(v)Let h be the height of vertex E, corresponding to the side BD in ΔBDE.
Let H be the height of vertex A, corresponding to the side BC in ΔABC. In (i), it was shown that ar (BDE) = 1/4 ar (ABC)
In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD).
⊥ ar (ΔBFE) = ar (ΔAFD)
(vi) Area (AFC) = area (AFD) + area (ADC)