Question 1204851
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Because segment AB is a diameter, triangle ABC is a right triangle with C = 90 degrees. This is due to Thale's Theorem. 
It's the special case of the Inscribed Angle Theorem.


Acute angle B of triangle ABC is 90-A = 90-32 = 58 degrees.
Since angle ABC = 58, so is angle DEB because of the Alternate Interior Angles Theorem. 
Note the parallel line markers on BC and DE.
Those are indeed parallel line markers and not vector symbols.


angleDEB + angleAED = 180
angleAED = 180 - angleDEB
angleAED = 180 - 58
angleAED = 122
This represents angle E of triangle AED.


Focus on triangle AED. 
This is isosceles due to radii ED = EA
The congruent base angles A and D are opposite the congruent sides ED and EA respectively.


For any triangle, the 3 inside angles always add to 180 degrees.
A+E+D = 180
x+122+x = 180
2x+122 = 180
2x = 180-122
2x = 58
x = 58/2
x = 29
Therefore, <font color=red>angle D is 29 degrees.</font>


Notice how interior angles A and D add to exterior angle DEB. 
Refer to the Remote Interior Angle Theorem for more info


The Remote Interior Angle Theorem is useful to help quickly prove the Inscribed Angle Theorem.


Edit
Greenestamps' answer is shockingly bad. 


It's very hand-wavy, and doesn't mention any theorems he used. 
Both are *really* bad practices in mathematics. 
Advice to students: Do NOT follow what Greenestamps did.
At least you know what not to do.


Not to mention there's a strange contradiction when he mentioned "arc BC is 64 degrees, arcs BC and CF are each 58 degrees". 
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