The vertices of a triangle ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC= ¼. Calculate the area of the angle ADE and compare it with the area of angle ABC. (Recall Theorem 6.2 and Theorem 6.6).

Given that: AD/AB = AE/AC = 1/4
Let AD = x, therefore AB = 4x as AD/AB = 1/4
Hence, BD = AB-AD = 4x-x = 3x
⇒ AD/DB = 1/3
Similarly,
AE/EC = 1/3
Hence, the points D and E divides the sides AB and AC respectively into 1:3.
Therefore,
Coordinates of point D = ((1×1+3×4)/(1+3), (1×5+3×6)/(1+3)) = (13/4, 23/4)
Coordinates of point E = ((1×7+3×4)/(1+3), (1×2+3×6)/(1+3)) = (19/4, 20/4)
Area of Triangle ADE
= 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
= 1/2[4((23/4)-(20/4))+13/4((20/4)-6)+19/4(6-(23/4))]
= 1/2[3- (13/4)+(19/16)]
= 1/2[(48-52+19)/16]
= 15/32 square units
Area of triangle ACB,
= 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
= 1/2[4(5-2)+1(2-6)+7(6-5)]
= 1/2[12-4+7]
= 15/2 square units
Now, (Area of triangle ADE)/(Area of triangle ABC) = (15/32)/(15/2) = 2/32 = 1/16

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