In Fig.9.33, ABC and BDE are two equilateraltriangles such that D is the mid-point of BC. If AEintersects BC at F, show that

(i) ar (BDE) =ar (ABC)

(ii) ar (BDE) =ar (BAE)

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

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

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

(vi) ar (FED) =ar (AFC)

[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 = 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 (ΔBDE) = 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) (i)

Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)

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

Area (ΔABD) = Area (ΔABE) (ii)

AD is the median in ΔABC

Area (ΔABD) = Area (ΔABC)

= Area (ΔBDE)

Area (ΔABD) = 2 Area (ΔBDE) (iii)

From (ii) and (iii), we obtain

2 ar (ΔBDE) = ar (ΔABE)

ar (ΔBDE) = ar (ΔABE)

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

ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)

Using equation (i), we obtain

ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC) ar (ΔABD) = ar (ΔBEC)

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