Question

State and prove Basic Proportionality Theorem or Thales Theorem.

Answer 1

To prove:

Thales Therorem i.e, If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Given:

A △ABC in which DE||BC and DE intersects AB at D and AC at E

R.T.P:-

$\frac{ad}{db} = \frac{ae}{ec}$

Construction:

Join BE and CD. Draw EN⊥AB and DM⊥BC

Proof:-

$area \: of \: triangle(ade) = \frac{1}{2} \times ad \times en$

$area\: of \: triangle(bde) = \frac{1}{2} \times bd \times en$

$\frac{area\: of \: tringle(ade) = \frac{1}{2} \times ad \times en }{area\: of \: triangle(bde) = \frac{1}{2} \times bd \times en}$

$\frac{area \: of \: triangle(ade)}{area\: of \: triangle(bde)} = \frac{ad}{bd} \: \: \: \: \: \: ....i)$

Similarly,

$area\: of \: triangle(ade) = \frac{1}{2} \times ab \times dm$

$area \: of \: triangle(cde) = \frac{1}{2} \times ec \times dm$

$\frac{area \: of \: triangle(ade)}{area \: of \: triangle(cde)} = \frac{ \frac{1}{2} \times ab \times dm}{ \frac{1}{2} \times ec \times dm }$

$\frac{area \: of \: triangle(ade)}{area \: of \: triangle(cde)} = \frac{ab}{ec}$

Therefore, △BDE and △CDE are on the same base DE and between the same parallels line DE and BC

ar(△BDE) =ar(△CDE)

Hence,

$\frac{area \: of \: triangle(ade)}{area \: of \: triangle(bde)} = \frac{ae}{ec} \: \: \: \: ......ii)$

From equation i) and ii)

$\frac{ad}{db} = \frac{ae}{ec}$

Answer 2

hi mate,

PROOF OF BPT

Given: In ΔABC, DE is parallel to BC

Line DE intersects sides AB and AC in points D and E respectively.

To Prove:

AD AE

----- = -----

DB AC

Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.

Proof:

Area of Triangle= ½ × base × height

In ΔADE and ΔBDE,

Ar(ADE) ½ ×AD×EF AD

----------- = ------------------ = ------ .....(1)

Ar(DBE) ½ ×DB×EF DB

In ΔADE and ΔCDE,

Ar(ADE) ½×AE×DG AE

------------ = --------------- = ------ ........(2)

Ar(ECD) ½×EC×DG EC

Note that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.

So, we can say that

Ar(ΔDBE)=Ar(ΔECD)

Therefore,

A(ΔADE) A(ΔADE)

------------- = ---------------

A(ΔBDE) A(ΔCDE)

Therefore,

AD AE

----- = -----

DB AC

Hence Proved.