Dividing a line segment into equal parts

Dividing a line segment into equal parts

This topic shows how to divide a given line segment into a number of equal parts with compass and straightedge or ruler. In our example, we divided it into five parts but it can be any number. By making use of a compass and straightedge construction, we do this without measuring the line.

Printable step-by-step instructions

The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not accessible.

Proof

The image below is the end drawing above with the AD, CB added and points labeled.

	Argument	Reason
We start by proving that AC, DB are parallel
1	AC = DB	By construction. See Copying a line segment for method and proof
2	AD = CB	By construction. Compass width for AD set from CB
3	ACBD is a parallelogram.	A quadrilateral with congruent opposite sides is a parallelogram.
4	AC, DB are parallel	Opposite sides of a parallelogram are parallel.
After that we prove that PE, QF are parallel
5	PQ = EF	There are drawn with same compass width
6	PQ, EF are parallel	From (4)
7	PQFE is a parallelogram.	A quadrilateral with one pair of opposite sides parallel and congruent is a parallelogram.
8	PE, GF are parallel	Opposite sides of a parallelogram are parallel.
Prove that triangle AQK is similar to and twice the size of APJ
9	∠APJ = ∠AQK	Corresponding angles. AB is a transversal across the parallels PE, QF
10	∠AJP = ∠AKQ	Corresponding angles. AB is a transversal across the parallels PE, QF
11	Triangles AQK, APJ are similar	AAA. ∠PAJ is common to both, and (9), (10) from Similar triangles test, angle-angle-angle theorem.
12	Triangles AQK is twice the size of APJ	AP = PQ. Both drawn with same compass width.
Prove that AJ = JK
13	AK is twice AJ	(11), (12). AQK is similar to, and twice the size of APJ. All sides of similar triangles are in the same proportion.
14	AJ = JK	From (13), J must be the midpoint of AK.
We have proved that the first two segments alongside the given line AB are congruent. We repeat steps 5-14 for each successive triangle. For instance we show that triangle ARL is similar to and three times APJ, and therefore AJ is one third AL. We move on till we have shown that all the segments along AB are congruent.
15	AJ = JK = KL = LM = MB	By applying the same steps to triangle AQK, ARL etc.
16	AB is divided into n equal parts.

- Q.E.D

This is the step-by-step, printable instruction is provided below to act as a guide in the absence of a computer.

	After you have done this	Your work ought to look like this
	Begin with a line segment AB that we will divide up into 5 (in this instance as already stated above) equal parts.
Step 1	From point A, draw a line segment at an angle to the given line, and roughly the same length. The precise length is not significant.
Step 2	Set the compasses on A, and set its width to a bit less than one fifth of the length of the new line.
Step 3	Step the compasses along the line, marking off 5 arcs. Label the last one C.
Step 4	With the compasses' width set to CB, draw an arc from A merely below it.
Step 5	With the compasses' width set to AC, draw an arc from B crossing the one drawn in step 4. This intersection is point D.
Step 6	Draw a line from D to B.
Step 7	With the use of the same compasses' width as used to step along AC, step the compasses from D along DB making 4 new arcs across the line
Step 8	Draw lines between the equivalent points along AC and DB.
Step 9	Done. The lines divide the given line segment AB in to 5 congruent parts.

Construction of angle 30°

This page illustrates the way to construct (draw) a 30 degree angle with compass and straightedge or ruler. It works by first constructing a rhombus and then a diagonal of that rhombus. With the use of the properties of a rhombus it can be illustrated that the angle produced has a measure of 30 degrees. See the proof below for detailed information on this.

Printable step-by-step instructions
Proof

This construction works by creating a rhombus. Its two diagonals form four 30-60-90 triangles.

The image below shows what the final drawing will look like with the red items added.

	Argument	Reason
1	Line segments PT, TR, RS, PS, TS are congruent (5 red lines)	All created with the same compass width.
2	PTRS is a rhombus.	A rhombus is a quadrilateral with four congruent sides.
3	Line segment AS is half the length of TS, and angle PAS is a right angle	Diagonals of a rhombus bisect each other at right angles.
4	Line segment AS is half the length of PS	PS is congruent to TS. See (1), (3)
5	Triangle ∆PAS is a 30-60-90 triangle.	∆PAS is a right triangle with two sides in the ratio 1:2. (third side would be √3 by Pythagoras).
6	Angle APS has a measure of 30°.	In any triangle, smallest angle is opposite shortest side.

- Q.E.D

This is the step-by-step, printable instruction to guide you in the absence of a computer

When you have done this	Your work ought to look like this
1. Draw a segment which will become one side of the angle. (leave out this step if you are given this line.) The exact length is not significant. Label it PQ. P will be the angle's vertex.
2. Put the compasses on P, and set its width to any suitable setting.
3. Draw an arc across PQ and up over above the point P. Label the point where it crosses PQ as point S.
4. Without altering the compasses' width, move the compasses to the point S. Draw a broad arc that crosses the first one and moves very well to the right. Label the point where the two arcs cross as point T.
5. Without altering the compasses' width, move the compasses to the point T, and draw an arc across the preceding arc, creating point R.
6. Draw a line from P to R.
Done. The angle QPR has a size of 30°

Construction of angle 45° Geometry construction using a compass and straightedge

This topic illustrates to way to construct (draw) a 45 degree angle with compass and straightedge or ruler. It works through first constructing an isosceles right triangle, which has interior angles of 45, 45 and 90 degrees. We make use of one of those 45 degree angles to obtain the result we require. Take a look at the proof below for more details.

Printable step-by-step instructions

This printable step-by-step instruction can be made use of for creating handouts or when a computer is not handy.

Proof

This construction works by constructing an isosceles right triangle, which is a 45-45-90 triangle. The image below is the end drawing to be obtained with the red items added.

	Argument	Reason
1	Line segment AB is perpendicular to PQ.	Constructed that way. Take a look at constructing the perpendicular bisector of a line.
2	Triangle APC is a right triangle	Angle ACP is 90° (from step 1)
3	Line segments CP,CA are congruent	Drawn with same compass width
4	Triangle ∆APC is isosceles.	CP = AC
5	Angle APC has a measure of 45°.	In isosceles triangle APC, base angles CPA and CAP are congruent. The third angle ACP is 90° and the interior angles of a triangle always add to 180. So both base angles CPA and CAP are 45°.

- Q.E.D

This is the step-by-step, printable instruction of how to construct angle 45°

After doing this	Your work should look like this
1. Draw a line segment which will become one side of the angle. (Skip this step if you are given this line.) The exact length is not important. Label it PQ. P will be the angle's vertex.
In the next 3 steps we produce the perpendicular bisector of PQ.
2. Set the compasses' width to be slightly over half the length of the line segment PQ.
3. With the compasses' point on P then Q, draw two arcs that cross above and below the line.
4. Draw a line between the two arc intersections. This is at right angles to PQ and bisects it (divides it in exactly half).
5. With the compasses' point on the intersection of PQ and the perpendicular not long drawn, set the compasses' width to P
6. Draw an arc across the perpendicular, producing the point C
7. Draw a line from P through C, and on a little more. The end of this line is point R
8. Done. The angle ∠QPR has a size of 45°

Constructing a 60° angle

This page illustrates how to construct (draw) a 60 degree angle with compass and straightedge or ruler. This construction works by producing an equilateral triangle. Recall that an equilateral triangle has all three interior angles 60°. We make use of one of those angles to obtain the desired 60 degree result.

Printable step-by-step instructions to guide you in the absence of a computer

Proof

This construction functions by producing an equilateral triangle. Recall that an equilateral triangle has all three interior angles 60°. The picture below is the final drawing above with the red items incorporated.

	Argument	Reason
1	Line segments AB, PB, PA are congruent	All drawn with the same compass width.
2	Triangle APB is an equilateral	Equilateral triangles are triangles with all three sides the same length.
3	Angle APB has a measure of 60°	All three interior angles of an equilateral triangle have a measure of 60°.

- Q.E.D.

After you have done the following	Your work ought to appear as shown below
1. Draw a line segment which will turn into one side of the angle. (Skip this step if you are given this line.) The exact length is not significant. Label it PQ. P will be the angle's vertex.
2. Set the compasses on P, and set its width to any suitable setting.
3. Draw an arc across PQ and up over above the point P.
4. Without altering the compasses' width, move the compasses to the point where the arc crosses PQ, and make an arc that crosses the first one.
5. Draw a line from P, through the intersection of the two arcs.
6. You are done at this point . The angle QPR has a size of 60°

View more on psalmfresh.blogspot.com for more educating and tricks article Thank you