Question

pls solve bpt theorem for me urgently in easy way

Answer 1

Given :- In triangle ABC, DE || BC

To Prove :- AD / BD = AE / CE

Construction :- Join DC, BE, draw DG Perpendicular to AC and EF Perpendicular to AB

Proof :-

ar (ADE) = AD × EF / 2 -----(1)

ar (BDE) = BD × EF / 2 ------(2)

On dividing equation 1 by 2, we get

ar ( ADE) / ar ( BDE) = AD / BD -------(3)


Similarly,

ar (ADE) / ar (DEC) = AE / EC ------(4)


But,

ar ( DEC) = ar ( BDE) ( They are on same base and between same parallel lines)

On putting this value in equation (4), we get

ar (ADE) / ar (BDE) = AE / EC -------(5)

On comparing equation (3) and (5), we get


AD/BD = AE / EC



Hence Proved...

Answer 2

Answer:-

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/DB = AE/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) / Ar(DBE)

= ½ ×AD×EF / ½ ×DB×EF

= AD/DB ......(1)

In ΔADE and ΔCDE,

=> Ar(ADE)/Ar(ECD)

= ½×AE×DG / ½×EC×DG

= AE/EC ........(2)

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(ΔBDE)

= A(ΔADE)/A(ΔCDE)

Therefore,

=> AD/DB = AE/AC

Hence Proved.

i hope it helps you.