Question

In the given figure, AB || DE and BD || EF. Prove that DC² = CF × AC.

Answer 1

Given: AB || DE, BD || EF

To prove: DC² = CF × AC

Proof: In ΔCDE and ΔCAB

∠D = ∠A   ...(Corresponding angles ∵ AB || DE)

∠E = ∠B

By AA test of similarity,

ΔCDE ∼ ΔCAB

`(CD)/(CA) = (CE)/(CB)`

⇒ `(CA)/(CD) = (CB)/(CE)`

⇒ `(CA)/(CD) - 1 = (CB)/(CE) - 1`

⇒ `(CA - CD)/(CD) = (CB - CE)/(CE)`

⇒ `(AD)/(CD) = (BE)/(CE)`   ...(1)

In ΔCFE and ΔCDB

∠F = ∠D   ...(Corresponding angles ∵ BD || EF)

∠E = ∠B

∴ ΔCFE ∼ ΔCDB

By AA similarity

`(CF)/(CD) = (CE)/(CB)`

⇒ `(CD)/(CF) = (CB)/(CE)`

⇒ `(CD)/(CF) - 1 = (CB)/(CE) - 1`

⇒ `(CD - CF)/(CF) = (CB - CE)/(CE)`

⇒  `(FD)/(CF) = (BE)/(CE)`   ...(2)

From equations (1) and (2)

`(AD)/(CD) = (FD)/(CF)`

`(AD)/(CD) + 1 = (FD)/(CF) + 1`

⇒ `(AD + CD)/(CD) = (FD + CF)/(CF)`

⇒ `(AC)/(CD) = (CD)/(CF)`

⇒ CD² = AC × CF

Hence proved.