Question

prove the BPT THEORM​

Answer 1

Given: In ΔABC, DE is parallel to BC

Line DE intersects sides AB and AC in points D and E respectively.

To Prove: ADBD=AECE

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)=12×AD×EF12×DB×EF=ADDB(1)

In ΔADE and ΔCDE,

Ar(ADE)Ar(ECD)=12×AE×DG12×EC×DG=AEEC(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,

ADBD=AECE

Hence Proved.

Plzz understand the diagram.

and mark it

Answer 2

plzz refer the attachment HOPE IT HELPS