Answer:
To prove that ∆ABC=8 ∆EFG, we need to use the concept of similarity of triangles and the ratio of their corresponding sides.
Given that ∆ABC and ∆EFG are similar triangles, we can write:
AB/EF = BC/FG = AC/EG = k (a constant)
Let's assume that AB = x, BC = y, and AC = z. Similarly, let EF = p, FG = q, and EG = r.
From the given information, we can write:
EF = AB/2 (since E is the midpoint of AB)
FG = BC/2 (since F is the midpoint of BC)
EG = AC/2 (since G is the midpoint of AC)
Substituting these values in the above equation, we get:
x/p = y/q = z/r = k
Now, let's consider the area of the triangles.
Area of ∆ABC = (1/2) * AB * BC * sin(∠BAC)
Area of ∆EFG = (1/2) * EF * FG * sin(∠EFG)
Using the values we have assumed earlier, we get:
Area of ∆ABC = (1/2) * x * y * sin(∠BAC)
Area of ∆EFG = (1/2) * (x/2) * (y/2) * sin(∠EFG)
Simplifying these expressions, we get:
Area of ∆ABC = (xy/2) * sin(∠BAC)
Area of ∆EFG = (xy/8) * sin(∠EFG)
Now, since the triangles are similar, their corresponding angles are equal. Therefore,
sin(∠BAC) / sin(∠EFG) = z/r
Substituting the value of k from earlier, we get:
sin(∠BAC) / sin(∠EFG) = 2k
Solving for sin(∠EFG), we get:
sin(∠EFG) = sin(∠BAC) / (2k)
Substituting this value in the expression for the area of ∆EFG, we get:
Area of ∆EFG = (xy/8) * (sin(∠BAC) / (2k))
Area of ∆EFG = (xy/16) * sin(∠BAC)
Now, substituting the value of the area of ∆ABC in this expression, we get:
Area of ∆EFG = (1/2) * Area of ∆ABC * (1/8)
Area of ∆EFG = (1/16) * Area of ∆ABC
Therefore, we have proved that ∆ABC=8 ∆EFG.
Explanation:
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