Question

state and prove basic proportionality theorem

Answer 1

in DE ll BC
prove that - AD/BD= AE /EC
in ADE = 1/2 ✖ b ✖ h
= 1/2 ✖ AD ✖ EN (because EN and. 1/2is cancelled )
in BED =1/2 ✖ DB ✖ EN
= AD /DB ...................1 equation
in AED = 1/2 ✖ AE ✖ EN
Area of CDE=1/2 ✖ CE ✖ EN
= AE /CE. ................2 equation
adding equation 1and 2 we get,
AD /DB =AE /CE
in ABC are of CED and Area of BED have a same base and same parallel
BED = CED
AD /BD= AE /EC

Answer 2

Answer:

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.