Answer:
Step-by-step explanation:
This is quite a doozy, so try and follow along. If you don't know your circle stuff well, it will be more difficult for you to understand how all this comes together.
First, we were given that angle CBD = 55. That means that angle CED is also 55 since they are both inscribed angles and they both intercept the same arc, arc CD. Angle BFC is congruent to angle 5 because they are vertical and vertical angles are congruent, so that means that triangle BFC is congruent to triangle EFD because of Angle-Angle-Side. Now that we got that out of the way, let's move on. We will come back to that in a bit.
Angle 1 intercepts arc BE and since angle 1 is an inscribed angle, it is half the measure of its intercepted arc. Therefore,
∠1 = 25° ; angle BCE also equals 25 degrees since they are both inscribed and they both intercept the same arc (again, arc BE).
If angle BCE is 25, and angle CBD is 55, then angle F, by the Triangle Angle Sum Theorem, must equal 180 - 25 - 55 = 100. And if Angle F = 100, then
∠5 = 100° . Angles 4 and 5 are supplementary, therefore,
∠4 = 80°
Since angle CBD was given as 55 and is an inscribed angle for arc CD, then arc CD is twice the measure of its inscribed angle and arc CD is 110 degrees. Angle 6, then, is one-half the difference of arc CD and arc BE so
∠6 = 30°
Angles 2 and 3 are congruent since they are both inscribed and they both intercept the same arc (arc DE); and since triangles BFC and EFD are congruent, by CPCTC, sides BF and EF are congruent...which means by the IsoscelesTriangle Theorem, angle 2 and angle BEC are congruent. 180 - 80 = 100 and divide that in half to get that
∠2 ≅ ∠3 = 50°
Lastly, angle 7 is half the measure of its intercepted arc (arc BE, given as 50), therefore,
∠7 = 25°
