Short Answer Type
Given: E is any point on median AD of a ΔABC.
To Prove: ar(ΔABE) = ar(ΔACE).
Proof: In ΔABC,
∵    AD is a median. ∴ ar(ΔABD) = ar(ΔACD)    ...(1)
∵ A median of a triangle divides it into two triangles of equal areas
In ΔEBC,
∵    ED is a median.
∴ ar(ΔEBD) = ar(ΔECD)    ...(2)
∵ A median of a triangle divides it into two triangles of equal areas Subtracting (2) from (1), we get ar(ΔABD) – ar(ΔEBD)
= ar(ΔACD) – ar(ΔECD)
⇒ ar(ΔABE) = ar(ΔACE).
Long Answer TypeD, E and F are respectively the midpoints of the sides BC, CA and AB of a ΔABC. Show that:
(i) BDEF is a parallelogram
In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar(ΔDOC) = ar(ΔAOB)
(ii) ar(ΔDCB) = ar(ΔACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint. From D and B, draw perpendiculars to AC.]
Short Answer Type
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