as height is 3 units and base is 6 units of triangle BCE then area =1/2*6*3 =9 units
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p and q are mid point of ac
hence pq || ac
and pq = 1/2 of ac
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Draw ΔPQR and mark the midpoints of PQ and PR as M and N respectively.
Consider ΔMNP and ΔPQR,
∠P = ∠P (common angle)
PN x 2 = PR
PM x 2 = PQ
If one angle of a triangle is equal to one angle of another triangle, and the length of the sides of the angle of the first is equal to the length of the sides of the angle of the second, then both triangles will be similar.
∴ ΔMNP ~ ΔPQR.
∴ ∠N = ∠R,
∠M = ∠Q,
and MN x 2 = QR.
∴ the length of the line joining the midpoints of any two sides of a triangle is half of that of the third side.
If ∠N = ∠R and ∠M = ∠Q, MN should be parallel to QR because PR and PQ cut MN and QR.
Now let's find the area of ΔBCE.
The distance between DC and BC is also the distance between DE and BC (BD), i. e., 3 units.
Because, on considering ΔBCE and ΔBCD, DE || BC and both triangles have the same base (BC). Two triangles, which have same base and lie on the same parallel lines, have same area and same altitude.
∴ If we draw an altitude from E to BC, its length is the same as BD = 3 units.
Area of ΔBCE
= 1/2 x BC x BD
= 1/2 x 6 x 3
= 1/2 x 18 = 9 square unit.
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