show that a line intersecting the two sides of a triangle and parallel to the third side of triangle divides the sides of the triangle in equal ratio
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In ΔABC, shown in the picture above, the line DE intersects AB and AC at points D and E respectively and DE || BC.
∴ ∠ADE = ∠ABC (corresponding angles)
∠AED = ∠ACB (,,)
∠DAE = ∠BAC (same angle)
∴ ΔADE and ΔABC are congruent
∴ AD/AB = AE/AC
=> AB/AD = AC/AE
=> (AD+DB)/AD = (AE+EC)/AE
=> 1+(DB/AD) = 1+(EC/AE)
=> DB/AD = EC/AE
=> AD/DB = AE/EC
=> AD:DB = AE:EC (Proved)
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