Can anyone help me with this problem? I think I probably miss some theorems.
In the figure angle $AED$ = $30$ deg, and the minor arc $AB, BC,$ and $CD$ are all equal. The degree measure of the minor arc $AD$ is
a) not enought information, b) $30$ deg, c) $105$ deg, d) $45$ deg, e) none of these.
2 Answers
$\begingroup$From the symmetry that $AB=CD$, we know that $\Delta BCE$ is isosceles and that lines $AD$ and $BC$ are parallel. We can add some angles to the original diagram:
The major arc $BD = 2\cdot105^\circ$, hence $AB=BC=CD=105^\circ$ and minor arc $AD=45^\circ$.
$\endgroup$ $\begingroup$A little bit more on how to prove this is as follows.
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
In the question angles BAC and ACD are inscribed angles.
If an angle is inscribe in a circle, then its measure is half the measure of its intercepted arc.
From this we conclude that angle BAC = (BC)/2 and angle ACD = (AD)/2.
Now from the exterior angle theorem we have.
Angle AED = abs[ angle BAC - angle ACD ]
Now if we substitute our previous equations into this exterior angle theorem we get.
Angle AED = abs [ (BC)/2 - (AD)/2]
The absolute value is there to make sure the angle is positive.
My final result is
which means that it depends on the radius r of the circle. So I don't think there is enough information to solve.
$\endgroup$ 0
