In the diagram shown, PC is tangent to the circle and PD is the angle bisector of \(\angle CPE\). If arc CD is \(70^\circ\), arc DE is \(30^\circ\), and \(\angle DQE = 40^\circ\), then determine arc AE, in degrees. 
+128407 Because chords CE and BD intersect within the circle, we can use the intersecting chords theorem to find minor arc BC
mDQE = (1/2) ( arc DE + arc BC )
40 = (1/2) ( 30 + arc BC)
80 = 30 + arc BC
50° = minor arc BC
And arc BAE =
360 - minor arc BC - minor arc CD - minor arc DE =
360 - 50 - 70 - 30 =
360 - ( 150) =
210°
And since PC is a tangent meeting the chord CE....the measure of angle PCE = (1/2) ( minor arc BC and arc BAE ) = (1/2) ( 50 + 210) = (1/2)(260) = 130° = mPCQ
And angle CQP = 40°
So angle CPQ = 180 - mPCQ - mCQP = 180 - 130 - 40 = 10°
And angle CPE is twice this = 20°
So....angle CEP = 180 - mPCE - mCPE = 180 - 130 - 20 = 30°
And because angle CEP is an inscribed angle, then minor arc CA is twice this = 60°
So
arc AE = 360 - minor arc CA - minor arc CD - minor arc DE =
360 - 60 - 70 - 30 =
360 - [ 60 + 70 + 30 ] =
360 - [ 160 ] =
200°
This does not not look to be possible....I think the problem is that minor arc DE is given as 30°....however....by the intersecting chords theorem, minor arc BC is computed to be greater than this
But..... minor arc DE looks to be much greater than minor arc BC (???)
