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Question
In the given figure, ABC is a triangle and AD is the median.
If E is the midpoint of the median AD, prove that: Area of ΔABC = 4 × Area of ΔABE
Solution
AD is the median of ΔABC.
Therefore it will divide ΔABC into two triangles of equal areas.
∴ Area(ΔABD) = Area(ΔACD) ….(i)
Similarly, ED is the median of ΔEBC.
∴ Area(ΔEBD) = Area(ΔECD) ….(ii)
Subtracting equation (ii) from (i), we have
Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD)
⇒ Area(ΔABE) = Area(ΔACE) ….(iii)
Since E is the mid-point of median AD,
AE = ED
Now,
ΔABE and ΔBED have equal bases and a common vertex B.
∴ Area(ΔABE) = Area(ΔBED) ….(iv)
From (i), (ii), (iii) and (iv), we get
Area(ΔABE) = A(ΔBED) = Area(ΔACE) = Area(ΔEDC) ….(v)
Now,
Area(ΔABC) = Area(ΔABE) + A(ΔBED) + Area(ΔACE) + Area(ΔEDC)
= 4 × Area(ΔABE). [From (v)]
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