| Use only your compass and straight edge when drawing a construction. No free-hand drawing! | | **Bisect a line segment** Note: This construction is also the construction for **Perpendicular Bisector of a Segment.**
| | **Given:** (a line segment) **Construction:** bisect . | | **STEPS:** **1. **Place your compass point on A and stretch the compass MORE THAN half way to point B (you may also stretch to point B). **2.** With this length, swing a large arc that will go above and below . **3.** Without changing the span on the compass, place the compass point on B and swing the arc again. The two arcs need to be extended sufficiently so they will intersect in two locations. **4. **Using your straightedge, connect the two points of intersection with a line or segment to locate point C which bisects the segment. | You may also see this construction done where only small portions of the arcs are shown both above and below the segment. The larger arcs, as shown here, are often easier to remember (since they form a "crayfish" looking creature) and they visually reinforce the circle concept needed for the proof of this construction.
| | The congruent triangle proof of this simple construction is quite lengthy. Here we go! | | **Proof of Construction: **Label the points of intersection of the arcs with the letters D and E. Draw segments These four segments will be congruent as they are the radii of two congruent circles. We can now show that there are four congruent triangles in this diagram giving us the congruent segments that will prove that the bisecting has occurred. This proof will use two sets of congruent triangles. **First **
** Triangles:** With we know ΔDAE and ΔDBE are congruent by SSS (they share side).
Now, ∠BDE ∠ADE since corresponding parts of Δs are (CPCTC). | **Second **
** Triangles:** With ∠BDE ∠ADE, and the shared side , ΔADC and ΔBDC are congruent by SAS. Now, by CPCTC. We now have bisecting since two congruent segments were formed. | **Perpendicular:**
We also have ∠DCA ∠DCB by CPCTC.
These ∠s form a linear pair making them supplementary. m∠DCA + m∠DCB = 180 (definition of supplementary). By substitution, m∠DCA + m∠DCB = 180, making m∠DCA = 90. ∠DCA is a right angle by definition, making since ⊥lines form rt.∠s. | | Notice that after this construction, the sides of quadrilateral ADBE are congruent making ADBE a rhombus. We know that the diagonals of a rhombus bisect each other and are also perpendicular, supporting our construction. | Also, keep in mind that all points on the perpendicular bisector of a segment are equidistant from the endpoints of the segment, which can be seen in this construction. **Given:** ∠ABC **Construction:** bisect ∠ABC.
| | **STEPS:** **1. **Place compass point on the vertex of the angle (point B). **2.** Stretch the compass to any length that will stay ON the angle. **3.** Swing an arc so the pencil crosses both sides (rays) of the given angle. You should now have two intersection points with the sides (rays) of the angle. **4. **Place the compass point on one of these new intersection points on the sides of the angle. If needed, stretch the compass to a sufficient length to place your pencil well into the interior of the angle. Stay between the sides (rays) of the angle. Place an arc in this interior (it is not necessary to cross the sides of the angle).
**5. **Without changing the span on the compass, place the point of the compass on the other intersection point on the side of the angle and make a similar arc. The two small arcs in the interior of the angle should be intersecting.
**6.** Connect the vertex of the angle (point B) to this intersection of the two small arcs. You now have two new angles of equal measure, with each being half of the original given angle. **Proof of Construction: **Label the points where the first arc intersects with the sides (rays) of the angle as E and F. The intersection of the two small arcs will be labeled D. Draw . By the construction, BE = BF and ED = FD (radii of the same circles). In addition, BD = BD. All of these equal length segments are also congruent, making ΔBED ΔBFD by SSS. Since corresponding parts of congruent triangles are congruent, ∠ABD ∠CBD, showing bisects ∠ABC. **NOTE: **The **re-posting of materials** (in part or whole) from this site to the Internet is **copyright violation** and is not considered "fair use" for educators. Please read the "Terms of Use".
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