Answer:
In ΔECD
<E = <C = <D
using angle sum property of triangle
<E + <C + <D = 180°
<E + <E + <E = 180° ( As <E = <C = <D )
3<E = 180
<E = 60°
NOW
<BCA + <ACE + <ECD = 180 (L. P. A.)
50° + <ACE + 60° =180
110 + <ACE = 180
<ACE = 70°
NOW as
AC = AE
So using property of triangle that angle opposite to equal sides are equal we get
<ACE = <AEC = 70°
In triangle ΔAEC
Now using Angle sum property of triangle
<ACE + <AEC + <CAE = 180°
70° + 70° + x = 180
140 + x = 180
x = 40°
Now in ΔABC
AB = AC
using property of triangle that angle opposite to equal sides are equal
So <ACB = <ABC = 50°
Now using angle sum property of triangle in ΔABC
<ACB + <ABC + <CAB = 180
50 + 50 + y = 180
100 + y = 180
y = 80°
so x = 40° and y = 80°
