I've to solve this simple exercise but i can't see how.
Problem: let $AB$ and $CD$ two equivalent ropes of one circle of centre $O$. Let $P$ and $Q$ two points that belong on the extentions of the previous ropes such that $BP$ and $DQ$ are equivalent. So show that the centre $O$ belong on the axis of the segment $PQ$ . rope of a circle

  • $\begingroup$ What is a "rope of a circle"? Can you provide a drawing? Even a snapshot of a hand drawing, with the points labelled, would suffice. $\endgroup$ – Brian Tung Aug 19 '15 at 18:33
  • $\begingroup$ @Brian Tung in the figure you can see the rope of a circle. I don't know the right english term $\endgroup$ – dario Aug 19 '15 at 18:41
  • $\begingroup$ Ahh, OK, they're called "chords." $\endgroup$ – Brian Tung Aug 19 '15 at 18:42
  • $\begingroup$ By "axis," do you mean "perpendicular bisector"? That is, the line that is perpendicular to $\overline{PQ}$ and intersects it at its midpoint? $\endgroup$ – Brian Tung Aug 19 '15 at 18:46
  • $\begingroup$ yes that's right $\endgroup$ – dario Aug 19 '15 at 20:57

First, show that $OP = OQ$; that is, both $P$ and $Q$ lie on a circle with center $O$:

  • $OB = OD$ (both on the same circle)
  • $BP = DQ$ (given)
  • $\triangle OAB \cong \triangle OCD$ (side-side-side, see below)

    • $OA = OC$ (both on the same circle)
    • $OB = OD$ (both on the same circle)
    • $AB = CD$ (given)
  • $m\angle OBA = m\angle ODC$ (corresponding angles of congruent triangles)

  • $m\angle OBP = m\angle ODQ$ (supplementary angles of equal angles)
  • $\triangle OBP = \triangle ODQ$ (side-angle-side)
  • $OP = OQ$ (corresponding sides of congruent triangles)

Now consider triangle $OPQ$. Let $R$ be the midpoint of $PQ$. Then

  • $PR = QR$ (midpoint)
  • $OP = OQ$ (proved above)
  • $\triangle OPQ$ is isosceles
  • $m\angle OPQ = m\angle OQP$ (base angles of isosceles triangle)
  • $\triangle OPR \cong \triangle OQR$ (side-angle-side)
  • $m\angle ORP = m\angle ORQ$ (corresponding angles of congruent triangles)
  • $m\angle ORP = m\angle ORQ = 90$ (congruent supplementary angles)

Therefore $\overline{OR}$ is the perpendicular bisector of $\overline{PQ}$.


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