Math, asked by jhanvi42, 10 months ago

prove BPT theorem................
....

.​

Answers

Answered by Anonymous
64

\huge\boxed{\fcolorbox{cyan}{Yellow}{ANSWER :-}}

PROOF OF BPT

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

For other example

check out the given attachment

✔️✔️PLEASE MARK ME AS BRAINLIEST✔️✔️

Attachments:
Answered by nilesh102
31

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.

Attachments:
Similar questions