Question

Given : AB || DE and BC || EF. Prove that :

1. `(AD)/(DG) = (CF)/(FG)`
2. ∆DFG ∼ ∆ACG

Answer 1

i. In ΔAGB, DE || AB, by Basic proportionality theorem

`(GD)/(DA) = (GE)/(EB)`   ...(1)

In ΔGBC, EF || BC, by Basic proportionality theorem,

`(GE)/(EB) = (GF)/(FC)`   ...(2)

From (1) and (2), we get

`(GD)/(DA) = (GF)/(FC)`

`(AD)/(DG) = (CF)/(FG)`

ii.

From (i), we have:

`(AD)/(DG) = (CF)/(FG)`

∠DGF = ∠AGC   ...(Common)

∴ ΔDFG ∼ ΔACG   ...(SAS similarity)