Basic proportionality Theorem or Thales Theorem class 10th

Imp points or results used in  Basic proportionality theorem or Thales theorem:-

(i) Area of Triangle = 1/2 base* height

Area of Triangle= 1/2 base*height
                    =1/2 BC*AD

(ii) Area of obtuse triangle = 

In this figure CM is height of the triangle and AB is the base
Area of triangle = 1/2 base* height
                   =1/2 AB* CM

(iii)Two triangles on the same base (or equal bases) and between the same parallels are equal in areas.(class 9th)

There are two triangle in this figure △CDA and △CDB
so,  ar(CDA)= ar(CDB)

Theorem 6.1

if a line is drawn parallel to one side of triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

 Proof

 Given:-  triangle ABC in which a line l parallel to side BC intersects other two sides AB and AC at D and E respectively. 

Prove that:- AD/ DB=AE/EC

Construction:- let us join BE and CD and then draw DG perpendicular on AC and EF perpendicular on AB.

 area of △ =1/2 base* height 

you may denote the area of △  as ar

so we can write area of △ADE  as ar(ADE).

so  ar(ADE)=1/2 AD*EF   .......(1)

         Also, ar(ADE)=1/2 AE*DG  .......(2)

similarly ar(BDE) =1/2 DB*EF   .......(3)  

                [Area of obtuse triangle result(ii)]

  

  and   ar(DEC)=1/2 EC*DG ............(4)

                [Area of obtuse triangle result(ii)]

[divide eq (1) by (3)]

 ar(ADE) / ar(BDE) =1/2 AD*EF / 1/2 DB*EF 

                      =AD/DB   ........(5)

and   [divide eq (2) by (4)]

ar(ADE)/ar(DEC)=1/2 AE*DG /1/2 EC*DG 

                 =AE/EC  ..........(6)

NOTE THAT:-

△BDE and △DEC are on the same base DE and between the same parellels BC and DE. [result(iii)]

so,         ar(BDE) = ar(DEC)   ........(7)

therefore, from eq (5),(6) and (7), we have:

AD/DB = AE/EC

Hence, the theorem is proved