Question

$\huge \pink \star{ \green{ \boxed{ \boxed{ \boxed{ \purple{ \mathfrak{question}}}}}}} \pink\star$
⭕Only maths brilliants
⭕Answer it.

"Proof of Thales Theorm"
​

Answer 1

$\huge\underline\underline\sf\blue{Statement:}$

If a line is drawn parallel to one side of a triangle, to interest the other two sides at indistinct points, the other two sides are divided in the same ratio.

$\huge\underline\underline\sf\blue{Given:}$ - In △ABC,DE∥BC

$\huge\underline\underline\sf\blue{To\:prove\:}$ AD/DB = AE/EC

$\huge\underline\underline\sf\blue{construction:}$

BE and CD are joined. EF⊥AB and DN⊥AL

 are drawn.

Proof:-

ar(△ADE) = 1/2×AD×EF

                  =1/2×AE×DN

ar(△BDE) = 1/2×BD×EF

ar(△CDE) = 1/2×EC×DN

ar(△ADE)/ar(△BDE) = 1/2×AD×EF/1/2×BD×EF   ........(1)

ar(△ADE)/ar(△CDE) = 1/2×AE×DN/1/2×EC×DN   ........(2)

But, ar(△BDE) = ar(△CDE)    ..............(3)

 

as they are on the same base DE and DE∥BC

 from (1), (2) and (3) we get AD/DB = AE/EC

Answer 2

$answer$

=>Given : In ∆ABC , DE || BC and intersects AB in D and AC in E.

=>Prove that : AD / DB = AE / EC

=>Construction : Join BC,CD and draw EF ┴ BA and DG ┴ CA.

=>Statements

Reasons

1) EF ┴ BA

1) Construction

2) EF is the height of ∆ADE and ∆DBE

2) Definition of perpendicular

3)Area(∆ADE) = (AD .EF)/2

3)Area = (Base .height)/2

4)Area(∆DBE) =(DB.EF)/2

4) Area = (Base .height)/2

5)(Area(∆ADE))/(Area(∆DBE)) = AD/DB

5) Divide (4) by (5)

6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC

6) Same as above

7) ∆DBE ~∆DEC

7) Both the ∆s are on the same base and

between the same || lines.

8) Area(∆DBE)=area(∆DEC)

8) If the two triangles are similar their

areas are equal

9) AD/DB =AE/EC

9) From (5) and (6) and (7)

=>In ∆ABC , if DE || BC and intersects AB in D and AC in E then

AD AE

---- = ------

DB EC

=>Proof on Thales theorem :

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.