Question

answer the question given with attachment ​

Answer 1

here is your answer-

Given,

<A=<CED

To prove,

∆CAB~∆CED

Proof:-

In∆CABand ∆CED

<A =<CED (given)

<C=<C ( common)

so, ∆CAB~∆CED (by AA)

AB/ED=BC/DC=AC/EC (by cpst)

so, BC/DC=AC/EC=3/2

also, AB/ED=3/2

9/x=3/2

x=6

I hope this will help you ☺☺☺