# simple exercise of euclidean geometry

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