Explanation:
an inscribed angle is an angle whose vertex lies on the circumference of a circle, while its two sides are chords of the same circle.
1.
we have to look for all vertexes on the circle circumference. from each one we have lines (chords of the circle) going to other points on the circle.
the angles at these vertexes are our inscribed angles.
at A.
the inscribed angle is EAD.
the corresponding arc is EFD.
at C.
the inscribed angle is ECF.
the corresponding arc is EF.
at D.
there are 3 lines defining 3 inscribed angles.
one inscribed angle is FDA.
the corresponding arc is FEA.
the second inscribed angle is FDE.
the corresponding arc is EF.
the third inscribed angle is EDA.
the corresponding arc is EA.
at F.
the inscribed angle is CFD.
the corresponding arc is CD.
at E.
there are 4 lines, but only 3 connect to the circle again, defining 3 inscribed angles.
one inscribed angle is AEC.
the corresponding arc is AC.
the second inscribed angle is CED.
the corresponding arc is CD.
the third inscribed angle is AED.
the corresponding arc is ACD.
2.
FDA and CFD.
why ?
because the triangle FBD must be isoceles (2 equally long legs = the radius of the circle), and so the triangle angles at F and D must be equal. which are the angles FDA and CFD.
3.
a.
the angle ECF is half of the angle EBF = 30/2 = 15°.
the arc angle EF is as defined 30°.
b.
the angle AED must be 90° (because the triangle AED is inscribed in a semicircle with its baseline being the diameter).
from that angle we know the part CED, which is again half of the angle CBD = 100/2 = 50°.
the angle AEC = angle AED - angle CED =
= 90 - 50 = 40°
the arc angle AC = complementary angle to arc angle CD (100°) =
= 180 - 100 = 80°
c.
maybe it is because it is late night, but I only can think of a multi-step solution. so, please try to follow :
as CF and AD are diameters, we have arc angles CDF and AEFD to be 180°.
arc angle CDF = angle CBD + angle FBD
180 = 100 + angle FBD
angle FBD = 80°
arc angle AEFD = angle FBD + angle EBF + angle ABE
180 = 80 + 30 + angle ABE
angle ABE = 70°
we also see that the triangle ABE is isoceles (same principle as with triangle FBD).
the triangle internal angles at A and E must be equal. and together with the angle at B (angle ABE) this must be 180°, as the sum of all 3 angles in a triangle must be 180°.
angle A = (180 - angle ABE)/2 = (180-70)/2 = 110/2 = 55°.
so, from the triangle ADE we know now 2 angles (A = 55°, E = 90° - see also b.). because of the 180° rule for triangle again
180 = 90 + 55 + angle ADE
angle ADE = 35°
arc angle AE = angle ABE = 70° (see above).