Question

Theorem 6.1 : prove that If a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same
ratio.​

Answer 1

⠀⠀⠀⠀⠀$\huge\underline\mathrm{Thales'\:Theorem-}$

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$\bold{\underline{\rm{\blue{Statement-}}}}$

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

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$\bold{\underline{\rm{\blue{Given-}}}}$

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• In ∆ABC, BC || DE

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$\bold{\underline{\rm{\blue{To\:prove-}}}}$

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• $\rm{\dfrac{AD}{BD}\:=\dfrac{AE}{EC}}$

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$\bold{\underline{\rm{\blue{Construction-}}}}$

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• Join BE and CD.
• Draw EF ⊥ AB and DG ⊥ AC

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$\bold{\underline{\rm{\blue{Proof-}}}}$

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We know that,

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$\large{\boxed{\rm{Area\:of\:\triangle=\dfrac{1}{2}\times{Base}\times{Height}}}}$

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$\rm{\dfrac{ar \triangle \: ADE}{ar \triangle \: BDE} = \dfrac{ \cancel{\frac{1}{2}} \times AD \times \cancel{EF}}{ \cancel{ \frac{1}{2}} \times {BD} \times \cancel{EF} } }$

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: $\implies$ $\rm{\dfrac{ar \triangle \: ADE}{ar \triangle \: BDE} = \dfrac{AD}{BD}}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀– eq (1)

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Now,

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$\rm{\dfrac{ar \triangle \: ADE}{ar \triangle \: CDE} = \dfrac{ \cancel{\frac{1}{2}} \times AE \times \cancel{DG}}{ \cancel{ \frac{1}{2}} \times {CE} \times \cancel{DG} } }$

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: $\implies$ $\rm{\dfrac{ar \triangle \: ADE}{ar \triangle \: CDE} = \dfrac{AE}{EC}}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀– eq (2)

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Also, we know that, triangles having same base and lying between same parallel lines are equal in area.

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$\therefore$ $\rm{Ar\:\triangle\:BDE\:=\:Ar\:\triangle\:CDE}$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀– eq (3)

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★ Using (1),(2) and (3), we get,

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$\large{\rm{\boxed{\red{\dfrac{AD}{BD}\:=\dfrac{AE}{EC}}}}}$

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Hence proved!