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Question
Given : AB || DE and BC || EF. Prove that :
- `(AD)/(DG) = (CF)/(FG)`
- ∆DFG ∼ ∆ACG
Solution
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)
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