Let BD = x , EC = y , AH = z Now , AG × AF = AH × AJ ⇒ 2 × 15 = z × (z + 7) ⇒ z2 + 7z - 30 = 0 ⇒ z = 3 Since, AC = AB ⇒ 16 = 3 + 7 + BJ ⇒ BJ = 6 Now , BJ × BH = BD × BE ⇒ 6 × 13 = x × (16 - y) ..... (i) Also , CE × CD = CF × CG ⇒ y (16 - x) = 1 × 14 ..... (ii) On solving equations (i) and (ii) we get x = 10 - √22 , y = 6 - √22 ∴ DE = 16 - x - y = 16 - (10 - √22) - (6 - √22) = 2 √22