prove the thales theorem
Answers
to one side of a triangle to intersect the other two sides in distinct points the other two sides are divided in the same ratio
in angle ADE and angle BDE
area of triangle ADE= 1/2× AD×NE
area of triangle BDE=1/2 × BD×NE
area
of tri ADE = AD/BDE
area of tri BDE
IN ANGLE ADE AND ANGLE CDE
AREA OF TRI ADE /AREA OF TRI CDE = 1/2×AE×DM/1/2 CE×DM
AREA OF TRI ADE/AREA OF TRI CDE =AE/CE HENCE AD/BD=AE/CE
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.
