In a, D and E are points on the sides AB and AC respectively. For each of the following cases show that :

(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.


(ii) AB = 5.6 cm, AD = 1.4 cm, AE = 7.2 cm and AC = 1.8 cm.


(iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.


(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm.


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(i) AB = 12 cm, AD = 8 cm, and AC = 18 cm.


DB=AB-AD


= 12-8


=4 cm


EC=AC-AE


= 18-12


= 6 cm


Now AD/DB=8/4=2


AE/EC=12/6=2


Thus DE divides side AB and AC of ABC in same ratio


Then by the converse of basic proportionality theorem.


(ii)



AB = 5.6 cm, AD = 1.4 cm, AE = 1.8 cm and AC = 7.2 cm


DB=AB-AD


DB=5.6-1.4


DB= 4.2 cm


And EC=AC-AE


EC= 7.2-1.8


EC=5.4


Now AD/DB=1.4/4.2=1/3


AE/EC=1.8/5.4=1/3


Thus DE divides side AB and AC of ABC in same ratio


Then by the converse of basic proportionality theorem.


(iii)



we have


AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm


AD=AB-DB


AD=10.8-4.5


AD= 6.3 cm


And EC=AC-AE


EC= 4.8-2.8


EC=2 cm


Now AD/DB=6.3/4.5=7/5


AE/EC=2.8/2=28/20=7/5


Thus DE divides side AB and AC of ABC in same ratio


Then by the converse of basic proportionality theorem.


(iv)



DEBC


We have,


AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm


Now AD/DB=5.7/9.5=57/95 =3/5


AE/EC=3.3/5.5=33/55=3/5


Thus DE divides side AB and AC of ABC in same ratio


Then by the converse of basic proportionality theorem.


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