In cyclic quadrilateral $ABCD$, let $AD$ and $CB$ meet at $P$. Let $E$ and $F$ be the midpoints of $DB$ and $CA$ respectively. Find $\angle PEF$ in terms of $\alpha, \beta, \gamma, \delta$, which are the angles of the cyclic quadrilateral.

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  • $\begingroup$ If the circle rotates, then the angles at $A,B,C,D$ stay constant. but it seems the angle $PEF$ will vary. $\endgroup$ – coffeemath Jul 26 '17 at 7:13
  • $\begingroup$ @coffeemath what do you mean by "If the circle rotates"? It seems that once you have a cyclic quadrilateral, all of the above is fixed? $\endgroup$ – Plato Jul 26 '17 at 7:22
  • $\begingroup$ Plato-- I guess my comment is not right, but maybe I was trying to express the ambiguity in definition of angle PEF noted in Artino's example below. $\endgroup$ – coffeemath Jul 26 '17 at 16:25

Angle $\angle PEF$, in general, does not depend only on the angles of the cyclic quadrilateral. See below an example, with two cyclic quadrilaterals having the same angles (they have parallel sides) but different values for $\angle PEF$. You also need to know the radius of the circle.

enter image description here


The radius of the circle is not enough: it is clear that one can scale the second figure so as to make two equal circles while keeping the same angles.

  • $\begingroup$ What do you think is enough to know? $\endgroup$ – Plato Jul 26 '17 at 18:21
  • $\begingroup$ One example could be knowing the eight angles, pairwise equal, formed by the diagonals with the sides. $\endgroup$ – Intelligenti pauca Jul 26 '17 at 19:25
  • $\begingroup$ Is it easy to calculate it from that? $\endgroup$ – Plato Jul 26 '17 at 19:45
  • $\begingroup$ I don't know. My guess is that such a formula would be very complicated, but I could be wrong of course. $\endgroup$ – Intelligenti pauca Jul 26 '17 at 20:21

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