Question

State basic proportionally theorem

Answer 1

Answer:

State basic proportionally theorem  : A line drawn parallel to

the third side of the triangle then the line divides the other two

sides in the same ratio.

In the picture given the diagram for basic proportionally

theorem

Given :

PQ ll BC

To Prove :

AP / PB = AQ / AC

Construction :

Draw QN perpendicular to AP

and PM perpendicular to AQ

Proof :

Consider triangle APQ and triangle PBQ

ar (APQ)/ar (PBQ) = 1/2 * AP * NQ/1/2 * PB * NQ = AP / PB ----- (1)

Consider triangle APQ and triangle PCQ

ar (APQ)/ar (PCQ) = 1/2 * AQ * PM/1/2 * QC * PM = AQ/AC --- (2)

Therefore, from (1) and (2)

ar (PBQ) = ar (PCQ) ----- (3)

Hence from (3)

AP / PB = AQ / AC

HENCE PROVED

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.