Using ruler and a pair of compasses only, construct a triangle PQR such that PQ = 6cm and angle PQR = 70 degrees. Construct perpendicular by setter of |PQ| and |QR|. Name the intersection of the mediatos D. Draw a circle with O as center and |OD| as radius.

• Draw a straight line segment $PQ$ with a length of $6 \text{ cm}$.
• Place the center of a protractor at point $Q$ and mark an angle of $70^\circ$ from $PQ$. Draw a ray $QX$ from $Q$ through this mark.
• With $Q$ as the center and a radius of $4 \text{ cm}$, draw an arc that intersects the ray $QX$. Label the intersection point $R$.
• Draw a straight line segment connecting points $P$ and $R$ to complete $\triangle PQR$.

• To construct the perpendicular bisector of $PQ$:
  • Place the compass needle at point $P$ and open it to a radius greater than half the length of $PQ$. Draw an arc above and an arc below $PQ$.
  • Without changing the compass radius, place the needle at point $Q$ and draw two more arcs that intersect the first two arcs.
  • Draw a straight line through the two points where the arcs intersect. This is the perpendicular bisector of $PQ$.
• To construct the perpendicular bisector of $QR$:
  • Place the compass needle at point $Q$ and open it to a radius greater than half the length of $QR$. Draw an arc above and an arc below $QR$.
  • Without changing the compass radius, place the needle at point $R$ and draw two more arcs that intersect the previous two arcs.
  • Draw a straight line through the two points where the arcs intersect. This is the perpendicular bisector of $QR$.
• Label the point where the two perpendicular bisectors intersect as $O$.

• Place the compass needle at point $O$ (the center) and open it to the length of $OP$ (the radius).
• Draw a circle. This circle should pass through points $P$, $Q$, and $R$.

The final answer is the completed geometric construction as described above.