Question

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

1. Prove that ΔADF ∼ ΔCEF.
2. Find AD if CE = 6 cm.
3. If DF is parallel to BC, find the 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)`

Hence, area of ΔADF : area of ΔABC = 25 : 64.