Question 8 plz answer fast and in detail, Thank you !

in Δ ABC & ΔADE

Δ ABC ≅ ΔADE as lines are extended in same ratio

AB/AD = BC/DE = AC/AE

=> AB/3AB = BC/DE = AC/3AC

=> 1/3 = BC/DE = 1/3

=> BC/DE = 1/3

Let say diagonals of BDEC cuts at O

inΔDOE & Δ BOC

∠BOC = ∠DOE (opposite angles)

∠CBO = ∠DEO as BC ║ DE

∠BCO = ∠EDO as BC ║ DE

ΔDOE ≅ Δ BOC

BC/DE = BO/EO = CO/DO

=> 1/3 = BO/EO = CO/OD

EO = 3BO DO = 3CO

EO : BO :: 3:1

DO : CO :: 3:1