Given: ∠GHE = ∠DFE = 90°, DH = 8, DF = 12, DG = 3x – 1 and DE = 4x + 2.

Find: the lengths of segments DG and DE.

In ΔDHG and ΔDFE,

⇒ ∠GHD = ∠DFE = 90°

⇒ ∠D = ∠D [Common]

Thus, ∆DHG ~ ∆DFE [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{DH}{DF} = \dfrac{DG}{DE} \\[1em] \Rightarrow \dfrac{8}{12} = \dfrac{(3x – 1)}{(4x + 2)} \\[1em] \Rightarrow 8(4x + 2) = 12(3x – 1) \\[1em] \Rightarrow 32x + 16 = 36x - 12 \\[1em] \Rightarrow 36x - 32x = 12 + 16 \\[1em] \Rightarrow 4x = 28 \\[1em] \Rightarrow x = 7.$

DG = 3x - 1 = 3(7) - 1 = 21 - 1 = 20,

DE = 4x + 2 = 4(7) + 2 = 28 + 2 = 30.

Hence, DG = 20 and DE = 30.