please help!

Dear Thais,

Thanks for your question. I am not sure though I understand it well.

If you meant to ask how to construct segment and angle bisectors using compass **and straightedge**,

then the answer is: the solutions are part of section 10 "Basic construction problems" of Kiselev's book

(please tell me if you don't have it).

However, if you are asking how to make such constructions by **compass alone**, then note that angle bisectors

are lines and simply cannot be drawn without straightedge. So, this question seems to make no sense.

In the contrary, the question about constructing segment bisectors using compass alone does make sense,

since such bisector is a point — the midpoint of the segment. So, this is a good question.

Let's try to solve it. According to Remark on p. 53 of the book, finding a solution to a construction problem should better

begin with the thought: "Suppose that what we want has already been constructed … "

So, suppose that C is the midpoint of a given segment AB. We can draw two circles of the *given* radius congruent to AB

and centered at A and C. These circles intersect at points D and E symmetric about the line AB. If only we could construct these

two points, we could recover C by intersecting two circles of the given radii DA=EA and centered at D and E.

Thus, here is the question: Can we construct D and E from A and B using only compass?

According to the advice on page 53, we should stare at the diagram showing the points A,B,C,D,E and try

to discover hidden relationships that would help us to invent a construction.

Actually there is an easier way out these days: Search the web! Here is a very nice applet I found this way,

which explains the solution.

Unfortunately it also spares the reader of the joy of discovery.

So, don't go there yet — Please answer my question first!

Sorry for being rather late with the reply,

Alexander

hello my name is lisa jane vaughn and i am a seinor at rubidoux high school in riverside california. i am doing a final project on segment bisectors and i need to know the history of a segment bisector like who created it, when it was created and why we use it today. if you could help me i would greatly appriciate it.

sincerly, lisa jane vaughn :D

Dear Lisa Jane,

I appreciate your interest in history.

In the 3rd century B.C. a Greek scholar **Euclid of Alexandria** wrote 13 books under the common title **The Elements,** apparently as a textbook for geometry students as well as a comprehensive exposition of basic mathematics of his time. The straightedge-and-compass constructions of angle and segment bisectors appear as Propositions 9 and 10 respectively in Book I of *The Elements.* For comparison: The last in Book I is Proposition 47 - *the Pythagorean Theorem.*

By now such drafting devices as straightedge and compass have been replaced with computer tools, but as recently as 50 years ago they were used by every engineer for producing technical drawings. Bisecting angles and segments were the most basic of the drafting techniques.

Nowadays, in countries with meaningful mathematics education, children learn how to bisect an angle or segment in early grades, as soon as it becomes safe to give them the compass. Then, from middle through high school they study geometry in the scope of several books of *Euclid's Elements.* By solving numerous challenging geometry problems, among which construction problems play a prominent role, they develop fine skills of creative, imaginative, and logical thinking, as well as intellectual curiosity and intuition. This is the main area of everyday application of angle and segment bisectors.

Here is an example of a moderately challenging construction problem which every high-school student graduating with flying colors should be able to handle using only straightedge and compass: **To construct one of the 50 stars of the American flag.**

Exercise 104: *Prove as a direct theorem that a point not lying on the perpendicular bisector of a segment is not equidistant from the endpoints of the segment; namely it is closer to that endpoint which lies on the same side of the bisector.*

Let there be a line segment $AB$ with perpendicular bisector $CD$. Select a point on one side of the plane about the perpendicular bisector, point $E$. Drop a perpendicular from point $E$ to line segment $AB$, the foot being point $F$. Draw slants from $E$ to $B$ and $E$ to $A$.

$52(2)$) If the perpendicular and some slants are drawn to a line from the same point outside this line, then if the feet of two slants are not the same distance away from the perpendicular, then the slant whose foot is farther away from the foot of the perpendicular is greater.

Slant $EB$ distance from $\perp$ is $FB$. Slant $EA$ distance from $\perp$ is $AC + CF$. $AC = CB$ $FB < CB$. Thus $E$ must be further from $A$ and closer to $B$, which is on the same side of the perpendicular bisector as $E$.

Exercise 105: *Prove as a direct theorem that any interior point of an angle which does no lie on the bisector is not equidistant from the sides of the angle.*

Let there be $\angle AOB$ with bisector $\overrightarrow{OC}$ and a point $E$ within the interior of $\angle AOC$. Drop perpendicular from $E$ to side $OA$, foot is point $G$. Drop perpendicular from $E$ to side $OB$, foot is point $H$. Drop perpendicular from $E$ to bisector $\overrightarrow{OC}$, and extend to intersect $OA$, at point $J$, and intersect $OB$, at point $K$. $JK$ intersects $\overrightarrow{OC}$ at point $F$.

Reflect the plane about $\overrightarrow{OC}$ such that side $OA$ identifies with side $OB$. Point $E$ takes point $E'$ within the interior of $\angle BOC$ and $G$ takes point $G'$ on side $OB$.

Construct rays $\overrightarrow{OE}$ and $\overrightarrow{OE'}$ from the vertex $O$. $EF = E'F$, $\angle OFE = \angle OFE'$, common side $OF$. $\S 40(1)$ SAS - $\triangle OFE = \triangle OFE'$, $OE = OE'$.

Construct slant $E'H$. $E'H > E'G$ as $\angle HG'E'$ is the greatest interior angle within $\triangle HG'E'$ ($\S 42$) and the side opposite to it is longest ($\S 45(2)$).

$\triangle OEH$ and $\triangle OE'H$ have two congruent sides $OE = OE'$ and $OH = OH$. $\angle E'OH$ is smaller than $\angle EOH$ due to side $OE'$ being on the interior of $\angle EOH$. By $\S 50(2)$ $EH > E'H$. But $E'H > E'G'$ and $E'G' = EG$. Thus $EH > EG$.

Exercise 106: *Prove that two perpendiculars to the sides of an angle erected at equal distances from the vertex meet on the bisector.*

Let there be $\angle AOB$ with bisector $\overrightarrow{OC}$. Choose a point on side $OA$, $D$. Mark segment $OD'$ along side $OB$ originating at point $O$, in the direction of $B$, and congruent to $OD$.

Erect perpendicular at point $D$ and continue it through bisector $\overrightarrow{OC}$, intersection point $E$. Erect perpendicular at point $D'$ and continue it through bisector $\overrightarrow{OC}$, intersection point $F$. $\S 40(2)$ ASA - $\triangle DOE = \triangle D'OF$, $\angle EOD = \angle EOD'$ ($\S15$), $OD = OD'$ by construction, and $\angle ODE = \angle OD'E = d$ ($\S 22$).

The sides $OE$ and $OF$ of these triangles are both along the bisector $\overrightarrow{OC}$, originate at $O$, and are congruent. Thus point $E$ and $F$ are identified and the same, and the perpendiculars meet on the bisector.

Exercise 107: *Prove that if A and A', and B and B' are two pairs of points symmetric about some line XY, then the four points A, A', B', B lie on the same circle.*

On one side of line $XY$ choose two points $A$ and $B$. Reflect the plane about $XY$, point $A$ takes position $A'$ and $B$ position $B'$. The respective pairs are on the same perpendicular to $XY$, $\S 37$. $XY$ is through the midpoints of said perpendiculars.

The center of the circle must be the same distance away from the four points, $\S 9$. $\S 56(2)$ - The points equidistant from the ends of either line segment of the pairs must be along the perpendicular at the midpoint of the respective segment. I.e. along $XY$. It is necesary to find a slant (radius) from $A$ to $XY$ and $B$ to $XY$ that are congruent, the circle's radius.

At midpoint, $D$, of $AB$ erect a perpendicular and extend it through $XY$, the intersection is $C$. Point $C$ is along $XY$ and equidistant from the respective pairs, but it is simultaneously along $CD$ and equidistant across pairs. $C$ is the center of the circle. Set the compass step to length $AC$ and construct the circle.

Exercise 108: *Find the geometric locus of vertices of isoceles triangles with a given base.*

$\S 33$) Isoceles triangle has two congruent sides. Base is side other than congruent sides. Vertex is angle formed by congruent sides.

Of the infinitely many triangles that can be constructed on one side of base $AC$, the one with a vertex point adjacent to the midpoint of $AC$ has the least altitude and is always contained on the interior by any with a greater altitude. Thus the midpoint of $AC$ is the only point in all of these triangles along the vertices.

Alternatively if the question is which points describe the vertex in any of these isoceles triangles.

$\S 36$) Perpendicular erected at the midpoint is also the bisector of the vertex.

The geometric locus of points is all points along the perpendicular erected at the midpoint of $AC$. Except the point of the foot of the perpendicular along $AC$.

Exercise 109: *Find the geometric locus of the vertices A of triangle ABC with the given base BC and such that angle B < angle C.*

Given $\angle B < \angle C$ it follows that $AB > AC$; $\S 45(2)$ - side opposite the greater angle is greater.

The geometric locus of vertices $A$ is all points where $AB > AC$. $\S 44(2)$ - angle opposite the greater side is greater.

Exercise 110: *Find the geometric locus of points equidistant from two given intersecting infinite straight lines.*

The point of intersection is the sole point in space occupied by both straight lines, $\S 4$.

In a given plane, a circle of arbitrary radius with its center point at this intersection, will be the locus of points equidistant from the intersection; $\S 9$.

If the question is asking only for points that lie on these two lines:

In any of these arbitrary radius circles, the points where the lines intersect the circle.

If the question is asking for points in the plane, which are not along the given lines, but equidistant from the lines:

The two given lines form four angles about the intersection point, which are vertical; $\S 26$. $\S 56(1)$) any point along the bisector of any angle is equidistant from the angle's sides. The geometric locus of points is all points along the bisectors of these 4 angles and those in the previously described circle(s) (not including those that are along the given lines).

If the locus of points is to be restricted to points that are equidistant from the lines, but this distance is the same for a given set:

Choose a point along the bisector of one of the angles, point $A$. This point is a distance $AZ$ from the lines / sides. There is a point $A'$ the same distance from the lines along the bisector of the angle vertical to this angle, a distance $OA$ from the vertex.

In the case that the angles formed by the intersection are right, there is another point the same distance from the lines in each of the other two angles formed by the intersection, $\S 22$.

In the other case, there will be two points along the bisectors of the other pair of vertical angles that are the same distance from the lines as points $A$ and $A'$; though not the same distance $OA$ from the vertex.

Exercise 111: *Find the geometric locus of points equidistant from three given infinite straight lines, intersecting pairwise.*

Pairwise intersection of three lines produces three points of intersection and a triangle. Depending on the figure, there may be a point on the interior of the triangle formed that is equidistant from two sides and also the base; thus equidistant from the three lines.

Such a point might also exist on the interior of this angle beyond the base of the triangle; i.e. equidistant from the 2 sides and the base. These points could exist for each intersection point taken as the vertex.