Explain basic proportionality theorm

Given : In {tex}\triangle A B C{/tex}, DE || BC and intersects AB in D and AC in E. 
Prove that : {tex}\frac{AD}{DB} = \frac{AE}{EC}{/tex}
Construction: Join BC, CD and draw EF {tex}\perp{/tex} BA and DG {tex}\perp{/tex} CA.  
Now from the given figure we have,
EF {tex}\perp{/tex} BA (Construction)
EF is the height of ∆ADE and ∆DBE    (Definition of perpendicular)
Area({tex}\triangle{/tex}ADE) ={tex}\frac{AD.EF}{2}{/tex}  .....(1)
Area({tex}\triangle{/tex}DBE) = {tex}\frac{DB.EF}{2}{/tex}   ....(2)
Divide the two equations we have
{tex}\frac{Area \triangle ADE}{Area \triangle DBE} = \frac{AD}{DB}{/tex}   .....(3)
{tex}\frac{Area \triangle ADE}{Area \triangle DEC} = \frac{AE}{EC}{/tex}   .....(4)
Therefore, {tex}\triangle \mathrm{DBE} \sim \triangle \mathrm{DEC}{/tex} (Both the ∆s are on the same base and between the same || lines).....(5)
Area({tex}\triangle{/tex}DBE) = Area({tex}\triangle{/tex}DEC) (If the two triangles are similar their areas are equal)
{tex}\frac{AD}{DB} = \frac{AE}{EC}{/tex} [from equation 3,4 and 5]
Hence proved.