In a triangle ABC, D and E are points on sides AB and AC respectively , such that DE is parallel to BC.If AD = 2.4cm, AE= 3.2cm , DE = 2cm and BC = 5cm , find BD and CE.

Same theorem should be applied here ...

Thales theorem....

According to the theorem..

AD/BD= AE/EC = DE/EC

= > AD/DB =DE/BC

=> 2.4/DB=2/5

=>  DB= 2.4*5/2

=> DB= 6cm

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Same theorem applied here too..

AE/EC = DE/BC

=>3.2/EC= DE/BC

=>EC= 3.2*5/2

=> EC= 8 cm

first prove triangle ABC similar triangle ade

angle A is common

angle B = angle D(BC//DE, corresponding angle are equal)

therefore triangle are similar

thus by proportional property of similar triangles

AB/AD = AC/AE =  BC/DE

or (AD+DB)/AD = BC/DE and (AE+EC)/AE =BC/DE

or 1+DB/AD = BC/DE  and 1+EC/AE = BC/DE

putting values of AD, BC, DE and AE

1+DB/2.4=5/2  and  1+EC/3.2 =5/2

1+DB/2.4 =2.5 and  1+EC/3.2 = 2.5

DB/2.4=2.5-1 and EC/3.2=2.5-1

DB/2.4=1.5  and  EC/3.2 =1.5

DB=1.5*2.4  and EC=1.5*3.2

DB=3.6 cm  and  EC=4.8 cm  ans

thanx......