Answer:
∠AED = 55°
Step-by-step explanation:
Two chords AB and CD are intersecting each other at E. Minor arc BD is = 70 and minor arc CA is = 180.
We have to find the measurement of ∠AED.
Now we know that ∠CEA = [tex]\frac{1}{2}(arcBD+arcCA)=\frac{1}{2}(70+180)=\frac{1}{2}(250)=125[/tex]
Since ∠CEA + ∠AED = 180° [ Supplementary angles ]
125° + ∠AED = 180
∠AED = 180 - 125 = 55°
Therefore measure of the angle AED = 55° is the answer.
