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A friend asked me two days ago showing this question.

He said that to me it would be very easy. He was wrong, I cannot solve it so far.

Prove that internal bisectors of angles in the four vertices of a quadrilateral determine by intersection four points which are common points to the same circle (see figure attached to prevent probable deficiencies in English).

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Piquito
  • 29,594
  • Denote the angles of the quadrilateral by A,B,C,D, and the angles in the small quadrilateral by P,Q,R,S. What is P in terms of A,B,C,D? What about R? What would be the relation between them if PQRS was cyclic? – almagest Mar 21 '16 at 17:26

1 Answers1

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$$ \angle DSC = \pi - \frac{1}{2} (\angle ADC +\angle BCD ) $$

$$ \angle AQB = \pi - \frac{1}{2} (\angle BAD +\angle ABC ) $$

Hence $$ \angle DSC +\angle AQB = \pi $$

HK Lee
  • 19,964
  • Amazing how math things can be either very easy or impossible, according to people involved. Thank you. – Piquito Mar 21 '16 at 18:07