Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
Let a ΔABC in which a line DE parallel to SC intersects AB at D and AC at E.
To prove DE divides the two sides in the same ratio.
i.e, `(AD)/(DB) = (AE)/(EC)`
Construction: Join BE, CD and draw EF ⊥ AB and DG ⊥ AC.
Proof: Here `(ar(ΔADE))/(ar(ΔBDE)) = (1/2 xx AD xx EF)/(1/2 xx DB xx EF)` .......[∵ Area of triangle = `1/2` × base × height]
= `(AD)/(DB)` .......(i)
Similarly, `(ar(ΔADE))/(ar(ΔDE)) = (1/2 xx AE xx GD)/(1/2 xx EC xx GD) = (AE)/(EG)` ......(ii)
Now, since, ΔBDE and ΔDEC lie between the same parallel DE and BC and on the same base DE.
So, ar(ΔBDE) = ar(ΔDEC) ......(iii)
From equations (i), (ii) and (iii),
`(AD)/(DB) = (AE)/(EC)`