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


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