If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio

#### Solution

Let ABC be a triangle in which a line DE is parallel to BC. It intersects the sides AB and AC at D and E respectively.

It has to be proved that,

`"AD"/"DB" = "AE"/"EC"`

Let us join BE and CD and draw perpendiculars DM and EN on AC and AB respectively.

Area of ΔADE `= 1/2 xx "Base" xx "Height"`

` = 1/2 xx "AD" xx "EN" = 1/2 xx "AE" xx "DM"`

Similarity, ar(ΔBDE) `= 1/2 xx "BD" xx "EN"`

ar (ΔDEC) `1/2 xx "EC" xx "DM"`

`("ar"(triangle"ADE"))/("ar"(triangle"BDE")) =(1/2xx"AD" xx "EN")/(1/2 xx "BD" xx "EN") = "AD"/"BD"`.....(1)

And, `("ar"(triangle"ADE"))/("ar"(triangle"DEC")) =(1/2xx"AE" xx "DM")/(1/2 xx "EC" xx "DM") = "AE"/"EC"`.....(2)

ΔBDE and ΔDEC are on the same base and between the same parallels.

∴ ar (ΔBDE) = ar (ΔDEC)

From (1) and (2), we obtain

`"AD"/"BD" ="AE"/"EC"`