Question

state and prove the basic proportionality theorem​

Answer 1

Answer:

If a line parallel to a side of a triangle intersect the remaining sides in two distinct point then the line divide the side in a same proportion

Step-by-step explanation:

poof ∆APQ and ∆PQB have equal height

A(∆APQ)=AP. 1.Area proportionate to base

A(∆PQB) PB

and A(∆APQ)=AQ. 2.area proportionate to base

A(∆PQC). QC

seg PQ is a common base of ∆PQR and ∆PQC. seg PQ || Seg BC

HENCE ∆ PQB and ∆PQC have equal height

A(∆PQB)= A(∆PQC). 3

A(∆APQ)=A(∆APQ). FROM 1,2 and3

A(∆PQB) A(∆PQC)

AP = AQ. FROM 1 and 2

PB. QC.

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.