(1)  \(\dfrac{1}{4}\sqrt{3}\)

Perhatikan \(\triangle \text{DAF}\) sama kaki, AD = AF (garis singgung yang sama panjangnya)

Garis AD dan AE sama-sama ditarik dari titik A, memiliki hubungan \(\text{AD}^2 = \text{AG} \times \text{AE}\)

\(\dfrac{\text{Luas segitiga DAF}}{\text{AG} \cdot \text{AE}} = \dfrac{\dfrac{1}{2} \cdot \text{AD} \cdot \text{AF} \cdot \sin 60^{\circ}}{\text{AD}^2}\)

\(\dfrac{\text{Luas segitiga DAF}}{\text{AG} \cdot \text{AE}} = \dfrac{\dfrac{1}{2} \cdot \text{AD} \cdot \text{AD} \cdot \sin 60^{\circ}}{\text{AD}^2}\)

\(\dfrac{\text{Luas segitiga DAF}}{\text{AG} \cdot \text{AE}} = \dfrac{\dfrac{1}{2} \cdot \cancel{\text{AD}^2} \cdot \dfrac{1}{2} \sqrt{3}}{\cancel{\text{AD}^2}}\)

\(\dfrac{\text{Luas segitiga DAF}}{\text{AG} \cdot \text{AE}} = \dfrac{1}{4} \sqrt{3}\)

(2)  6 cm

Perhatikan bahwa BD = BE dan CF = CE

BC = BE + EC

BC = BD + FC

BC = 4 + 2

BC = 6 cm

(3)  \((\sqrt{33}\:-\:3)\) cm

Perhatikan segitiga ABC, gunakan aturan cosinus.

Panjang AD = AF = \(x\) cm

\(\text{BC}^2 = \text{AB}^2 + \text{AC}^2 \:-\:2 \text{AB} \cdot \text{AC} \cdot \cos 60^{\circ}\)

\(6^2 = (x + 4)^2 + (x + 2)^2 \:-\:\cancel{2} \cdot (x + 4)(x + 2) \cdot \dfrac{1}{\cancel{2}}\)

\(36 = x^2 + 8x + 16 + x^2 + 4x + 4 \:-\:(x^2 + 6x + 8) \)

\(36 = x^2 + 8x + 16 + x^2 + 4x + 4 \:-\:x^2 \:-\:6x \:-\:8\)

\(36 = x^2 + 6x + 12\)

\(0 = x^2 + 6x \:-\:24\)

\(x = \dfrac{-b \pm \sqrt{b^2 \:-\:4ac}}{2a}\)

\(x = \dfrac{-6 \pm \sqrt{6^2 \:-\:4(1)(-24)}}{2(1)}\)

\(x = \dfrac{-6 \pm \sqrt{36 + 96}}{2}\)

\(x = \dfrac{-6 + \sqrt{132}}{2}\)

\(x = \dfrac{-6 + 2\sqrt{33}}{2}\)

\(x = -3 + \sqrt{33}\text{ cm}\)

\(x = (\sqrt{33}\:-\:3)\text{ cm}\)