Theorem 6.1

Dwnload pdf of this sollution from here

                     DOWNLOAD PDF                

theorem 6.1: If line is drawn parallel 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

Solution : - 

Given : In triangle ABC, BC||DE

To prove : AD/BD=AE/CE

Construction : Join BE and CD and

perpendicular of AB to E and AC to D

Proof : 

area of triangle ADE = 1/2 Base × height

ar(ADE) = 1/2×AD×EF ...(1)

area of triangle BDE = 1/2 Base × height

ar(BDE) = 1/2×BD×EF ...(2)

Now, From (1)/(2) we get

ar(ADE)= 1/2×AD×EF

ar(BDE)= 1/2×BD×EF

i.e.,

= AD/BD ...(3)

Similarly,

=AE/CE ...(4)

then, we know that two triangle between same base and same parallel DE and BC are equal in area

i.e., ar(BDE) = ar(CED) ...(5)

therefore, from (3),(4)&(5) we have

AD/BD = AE/CE

Hence, proved