A triangle is formed by joining the coordinates B(0,0), C(8,0) and A(4,8). DE is joined where D is the mid-point of AB, and E is the mid-point of AC. DC and BE is joined to intersect at F.
Then, Area of triangle DFB = (x) $\times$ Area of triangle EFC, where x is an integer. The value of x is __.

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Solution

Since, D is the mid-point of AB, and E is the mid-point of AC in triangle ABC. From mid-point theorem, line joining the mid points of two sides is parallel to third side and is equal to half of it.

DC $∥$ BC

Since, triangles between the same base and same parallels have equal area.
Thus, Area of $Δ$ DCB = Area of $Δ$ BEC.

$\Rightarrow$ Area of $Δ$ DCB- Area of $Δ$ BFC = Area of $Δ$ BEC- Area of $Δ$ BFC

$\Rightarrow$ Area of $Δ$ DFB = Area of $Δ$ EFC

$\Rightarrow$ Area of $Δ$ DFB = (1) $\times$ Area of $Δ$ EFC

Hence, value of x = 1

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A triangle is formed by joining the coordinates B(0,0), C(8,0) and A(4,8). DE is joined where D is the mid-point of AB, and E is the mid-point of AC. DC and BE is joined to intersect at F. Then, Area of triangle DFB = (x) × Area of triangle EFC, where x is an integer. The value of x is __.

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