The vertices of a Δ 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 
Calculate the area of theΔ ADE and compare it with the area of Δ ABC (Recall Theorem 6.2 and Theorem 6.6)
Given that,
= 
= 
Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3
Coordinates of point D = (
)
= (
, 
)
Coordinates of point E = (
, 
)
= (
, 
)
Area of the triangle = 
[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 - y2)
= 
[4 (
– 
) + 
(
- 6) + 
(6 – 
)]
= 
[3 – 
+ 
]
= 
[
]
= 15/32 square units
Area of triangle ABC = [4 (5 – 2) + 1 (2 - 6) + 7 (6 – 5)] /2
= (12 – 4 + 7)/ 2
= 15/2 square units
Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16
We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel tothe third side of the triangle.
These two triangles so formed (here ΔADE and ΔABC) will be similar to each other.
Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles
And, ratio between the areas of ΔADE and ΔABC = (1/4)2
= 1/16