Question

state and prove fundamental theorem of proportionality.

Answer 1

Here's the answer
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 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.

The BPT also has a converse which states, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

(Note: A converse of any theorem is just a reverse of the original theorem, just like we have active and passive voices in English.)

i hope it helps you.