consider a triangle abc. The sides ab and ac are extended to points d and e, respectively, such that ad = 3ab and ae = 3ac. Then one diagonal of bdec divides the other diagonal in what ratio ?
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Answer:
3:1
Step-by-step explanation:
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
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