Circle $P$ has diameter $CD$. Point $B$ is on the circle such that $m\angle BPC=30$ degrees. Point $A$ is on the circle such that $AD$ is parallel to $PB$.

What is the degree measure of arc $ABC$?

I am stuck on how to draw this diagram and go about solving the problem. Do realise the answer must be $60$ degrees, but don't understand how. Any help will be appreciated.

  • $\begingroup$ Please type your question in MathJax instead of using an image. $\endgroup$ – Toby Mak Feb 19 '20 at 8:51
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    $\begingroup$ What are you stuck on when drawing the diagram? $\endgroup$ – lioness99a Feb 19 '20 at 8:57
  • $\begingroup$ @Tumul Kumar Easy to see that $\measuredangle APC=2\cdot30^{\circ}=60^{\circ}.$ $\endgroup$ – Michael Rozenberg Feb 19 '20 at 8:58
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    $\begingroup$ @MichaelRozenberg how did you obtain that? $\endgroup$ – Tumul Kumar Feb 19 '20 at 9:00
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    $\begingroup$ Once you draw the diagram, find $\angle ADP$, $\angle PAD$, and $\angle APD$. If you have any more questions please comment below. $\endgroup$ – Toby Mak Feb 19 '20 at 9:03

This is my attempt at the drawing:

enter image description here

From the information in the comments, we find that $\angle PAD = \angle ADP = 30º$ ($PA=PD$ since they are both radii, so $\Delta PAD$ is isosceles. Therefore $\angle PAD = 120º$ which gives $\angle APC = 60º$.

This can also be found using the exterior angle theorem, which in this case, means that $\angle PAD + \angle ADP = \angle APC$.

Just to clarify, 'degree measure' means the angle of rotation from one side to the other side, not the angle in between the arc. This is further explained in this YouTube video.


This is my attempt on the drawing.

  • $\begingroup$ Sorry for wrong angle nomenclature, just flip the image in your head ^^ $\endgroup$ – Matthew Feb 19 '20 at 9:11
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    $\begingroup$ If I'm not wrong, the angle to be measured should be $\angle APC$. $\endgroup$ – Toby Mak Feb 19 '20 at 9:13
  • $\begingroup$ Correct, sorry for that. $\endgroup$ – Matthew Feb 19 '20 at 9:14

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