According to Thales theorem, if in a given Triangle a line is drawn parallel to any of the sides of the Triangle so that the other two sides intersect at some distinct point then it divides the two sides in the same ratio.

PROOF

Construction: ABC is a triangle; DE is a line parallel to BC and intersecting AB at D and AC at E, i.e. DE || BC.

Join C to D and B to E. Draw EM $\bot$ AB and DN $\bot$ AC.

We need to prove that AD/DB = AE/EC

Proof:

Area of a triangle, ADE = $1/2$ $\times$ AD $\times$ EM

Similarly,

Ar(BDE) = $1/2$ $\times$ DB $\times$ EM

Ar(ADE) = $1/2$ $\times$ AE $\times$ DN

Ar(DEC) = $1/2$ $\times$ EC $\times$ DN

Hence,

Ar(ADE)/Ar(BDE) = $1/2$ $\times$ AD $\times$ EM / $1/2$ $\times$ DB $\times$ EM = AD/DB

Similarly,

Ar(ADE)/Ar(DEC) = AE/EC

Triangles DEC and BDE are on the same base, i.e. DE and between same parallels DE and BC.

Hence,

Ar(BDE) = Ar(DEC)

From the above equations, we can say that

AD/DB = AE/EC.

Hence, proved.

Previous Topic

Back to Lesson

Next Topic