Question

In the given figure DE||BC if ab=7.2cm, ad =1.8cm then find ae:ec

Answer 1

Answer: 1 : 5

Step-by-step explanation:

Given : DE ║ BC ,

A is the intersection point of the Line segments CE and BD,

We have to find : AE : EC

Since, In triangles EAC and CAB,

By the Alternative interior angle theorem,

$\angle AED\cong \angles ACB$    

And, $\angle ADE\cong \angles ABC$

By AA similarity postulate,

$\triangle EAC\cong \triangle CAB$

By the property of similar triangles,

$\frac{DA}{BA}=\frac{AE}{AC}$

⇒ $1+\frac{DA}{BA}=1+\frac{AE}{AC}$

⇒ $\frac{BA+DA}{BA}=\frac{AC+AE}{AC}$

⇒ $\frac{7.2+1.8}{7.2}=\frac{CE}{AC}$

⇒ $\frac{9}{7.2}=\frac{CE}{AC}$

⇒ $\frac{5}{4}=\frac{CE}{AC}$ -------(1)

⇒ $\frac{5}{4}-1=\frac{CE}{AC}-1$

⇒ $\frac{5-4}{4}=\frac{CE-AC}{AC}$

⇒ $\frac{1}{4}=\frac{AE}{AC}$ -------(2)

Dividing equation (2) by equation (1),

$\frac{1}{5} = \frac{AE}{CE}$

Answer 2

Answer:

Step-by-step explanation:

DE||BC

AB=7.2 cm,AD=1.8cm

AD/AB=1.8/7.2

=1/4

So,AD=1x and BD=3x

AD/BD=AE/EC (by bpt)

So,AE/EC=1/3