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Question
In the given figure, ABC is a triangle. DE is parallel to BC and `"AD"/"DB" = (3)/(2)`.
(i) Determine the ratios `"AD"/"AB","DE"/"BC"`.
(ii) Prove that ΔDEF is similar to ΔCBF.
Hence, find `"EF"/"FB"`.
(iii) What is the ratio of the areas of ΔDEF and ΔBFC?
Solution
(i) Given
DE || BC
and `"AD"/"DB" = (3)/(2)`
In ΔADE and ΔABC,
∠A = ∠A, ...(Common Angles)
∠D = ∠B ...(Corresponding Angles)
∴ ΔADE ∼ ΔABC ...(by A.A. criterion)
∴ `"AD"/"AB" = "AE"/"AC" = "DE"/"BC"`
Now `"AD"/"AB" = "AD"/("AD"+"BC")`
= `(3)/(3 + 2) = (3)/(5)`
∴ `"AD"/"AB" = (3)/(5) = "DE"/"BC"`.
(ii) In ΔDEF and ΔCBF,
∠FDE = ∠FCB ...(Alternate Angle)
∠DFE = ∠BFC ...(Vertically Opposite Angle)
∴ ΔDEF ∼ ΔCBF ...(by A.A. criterion)
Hence proved.
`"EF"/"FB" = "DE"/"BC" = (3)/(5)`
∴ `"EF"/"FB" = (3)/(5)`.
(iii) `"Area of ΔDFE"/"Area of ΔCBF" = "EF"^2/"FB"^2 = 3^2/5^2 = (9)/(25)`.
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