Solve this: AC = CE and AB = BD. Find x$°$.

(A) 18$°$
(B) 10$°$
(C) 8$°$
(D) 28$°$

Dear Student,

Please find below the solution to the asked query:

Given : AC =  CE , So from base angle theorem in $∆$ ACE we get :

$\angle$ CAE =  $\angle$ CEA ,

$\angle$ CAE  = 40$°$                                     ( From given diagram we know : $\angle$ CEA  = 40$°$ )

Also given : AB =  BD , So from base angle theorem in $∆$ ABD we get :

$\angle$ BAD =  $\angle$ BDA ,

$\angle$ BAD  = 40$°$                                     ( From given diagram we know : $\angle$ BDA  = 40$°$ )

Now from angle sum property we triangle we get in $∆$ ADE : 

$\angle$ ADE + $\angle$ AED + $\angle$ DAE  = 180$°$ ,

40$°$ + 40$°$+ $\angle$ DAE  = 180$°$ ,    ( From given diagram $\angle$ ADE = $\angle$ AED = 40$°$ )

 $\angle$ DAE  = 100$°$ ,

 $\angle$ CAE + $\angle$ BAD  + $\angle$ BAC  = 100$°$ ,   ( From given diagram :  $\angle$ DAE  =  $\angle$ CAE + $\angle$ BAD  + $\angle$ BAC )

40$°$  + 40$°$ +  x = 100$°$ ,   ( Substitute from given diagram and as we calculated )

x = 20$°$                   ( Ans ) 

But our solution do not match with any of the given options so kindly recheck that information and get back to us so that we could help you precisely .

Regards