Question

In ∆ABC, seg DE || side BC. If DB = 8 cm, AD = 12 cm, EC = 10 cm then find AE and AC​

Answer 1

Triangle ADE and ABC are similar

so,

AD/AB=AE/AC

AD/(AD+DB)= AE/AC

12/(12+8) = AE/(AE+EC)

12/20 = AE/ (AE+10)

NOW DIVIDING 12 AND 20 BY 4

3/5= AE/(AE+10)

now cross multiply

3(AE+10) = 5AE

3AE +30 = 5AE

30= 5AE-3AE

30=2 AE

30/2 = AE

AE=15

SO AC = AE+EC = 15+10= 25

Answer 2

The basic $proportionality$ theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

It is given that AD=12 cm, DB = 8 cm , EC =10 cm

AB= AD + DB = 12 cm + 8 cm =20 cm

Let's use the basic$proportionality$ theorem,

AD /AB =AE / AC =DE / BC

AD /AB =AE /AC

$12/20$ =AE / ( AE + EC)

3/5 = AE /( AE + 10 )

3 ( AE + 10 ) = 5 AE

3 AE + 30 = 5 AE

5 AE - 3 AE = 30

2 AE = 30

AE = 15

∴AC = AE + EC = 15 + 10 = 25

Ans :-  AE is 15 cm ,  and AC is 25 cm