Discussion

Explanation:

Let A(∆ADE) = 1, hence A(DEGF) = A(FGCB) = 1

∴ A(∆AFG) = 2 and A(∆ABC) = 3

In the figure, DE || FG || BC, hence ∆ADE is similar to ∆AFG is similar to ∆ABC

Since, the 3 triangles are similar, the ratio of their areas will be same as the ratio of squares of their sides.

⇒ A(∆ADE) : A(∆AFG) : A(∆ABC) = 1 : 2 : 3 = AD² : AF² : AB²

⇒ AD : AF : AB = 1 : √2 : √3

Now, DF = AF - AD = √2 - 1
and, FB = AB - AF = √3 - √2

∴ AD : DF : FB = 1 : (√2 - 1) : (√3 - √2)

Hence, option (d).

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