Question

In the given figure ABC and CEF are two triangles where BA is parallel to CE and AF: AC = 5: 8.

(i) Prove that ΔADF ∼ ΔCEF
(ii) Find AD if CE = 6 cm
(iii) If DF is parallel to BC find area of ΔADF: area of ΔABC.

Answer 1

(i) In ΔADF and ΔCFE
∠DAF = ∠FCE   ...(alternate angles)
∠AFD = ∠CEF  ...(vertically opp. angles)
∠ADF = ∠CEF
∴ ΔADF ∼ ΔCEF  ...(by A.A.)
Hence Proved.
(ii) ΔADF ∼ ΔCEF 
∴ `"AD"/"CE" = "AF"/"FC"`
FC = AC - AF
= 8 - 5 = 3
∴ `"AD"/(6) = (5)/(3)`
⇒ AD = 10 cm.
(iii) DF | | BC ∴ ΔADF ∼ ΔABC
∵ ∠D = ∠B and ∠F = ∠C.
∴ `"Ar. of ΔADF"/"Ar. of ΔABC" = "AF"^2/"AC"^2`
= `(5/8)^2 = (25)/(64)`.