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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.

Guest Dec 26, 2018

#1**+1 **

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 (???)

CPhill Dec 26, 2018