Dwnload pdf of this sollution from here DOWNLOAD PDF Theorem 6.2 : If a line divides any two sides of a triangle in a same ratio then and it is parallel to the third side. Sollution : - Given =》In triangle ABC, AB/BD=AE/CE To prove =》DE || BC Construction =》Draw DE' || BC Proof, Since DE' || BC (by const.) by theorem 6.1, AD/BD = AE'/CE' ...(1) But, given that AD/BD = AE/CE ...(2) From (1) & (2) we have, AE'/CE' = AE/CE ...(3) Now, AE' + CE' = AC ...(a) AE + EC = AC ...(b) Then, add 1 to eq(3) we get --> AE'/CE' + 1 = AE/CE + 1 -->AE'+CE'/CE' = AE+CE/CE From eq(a) & (b) we have, AC/CE' = AC/CE 1/CE' = 1/CE or, CE' = CE From above we can say that point A and E' are consideing therefore, DE' || BC i.e., DE || BC Hence, proved
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