find delta, maybe by congruent triangles? Hey there: I think this is a rather short and easy question for you. Can anyone either way please give me a hint? Would be very lovely!
I found out about almost all angles in this triangle. In my solution sheet it says that $\delta$ should be $10^\circ$. But how can I explain it or prove it? I don't have a clue how to find $\beta$ or $\delta$. Can anyone please give me a hint?
Hint: Angle subtended by a CHORD on the circumference is equal. (Given the fact that $w$ is the Angular bisector).



Notice here: $ \angle EBC=\angle EDC=70^0$. Now you can go on.
Given: 
A
C
‾
AC
  and 
B
D
‾
BD
  bisect each other.

Prove: 
B
C
‾
≅
A
D
‾
BC
 ≅ 
AD
 .

Note: quadrilateral properties are not permitted in this proof.
Step    Statement   Reason
1
A
C
‾
AC
  and 
B
D
‾
BD
  bisect each other
Given
try 
Type of Statement
A
B
C
D
E
Note: 
A
C
‾
AC
  and 
B
D
‾
BD
  are segments.

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