Question

prove thales theorem

Answer 1

STATEMENT:IF A LINE IS DRAWN PARALLEL TO SIDE OF A TRIANGLE TO INTERSECT THE OTHER TWO SIDES IN DISTINCT POINTS THEN THE OTHER TWO SIDES ARE DIVIDE IN THE SAME RATIO.

GIVEN:In Δ ABC DE is parallel to BC which intersects sides AB and AC at D and E respectively

RTP: AD / DB = AE / EC

CONSTRUCTION:Join B,E and C,D and then draw  DM is perpendicular to AC and EN perpendicular to AB

PROOF: Area of ΔADB = 1/2*AD*EN
             AREA of ΔBDE = 1/2*BD*EN

so, ar(ΔADB) / ar(ΔBDE) = 1/2*AD*EN / 1/2*BD*EN = AD/BD -------(1)
    
              area of ΔADE = 1/2*AE*DM
              area of ΔCDE = 1/2*EN*DM

SO,ar(ΔADE) / ar(ΔCDE) = 1/2*AE*DM / 1/2*EN*DM = AE/EN ----------(2)

we observe that ΔBDE and ΔCDE are on the same base DE and between the same parallel BC and DE.

so ar(ΔBDE) = ar(ΔCDE) -------(3)

from (1),(2),(3) we know

AD/DB = AE/EC

HENCE PROVED

Answer 2

Answer:-

Thales Theorem Is Also known As BPT Theorem

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.