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"`
