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Question 5 (i)
In the following 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 ar(BDE)=14ar(ABC)

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Solution

(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=12BC 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)=14ar(ABC)

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