But, intuitively, you should see that thearc is all of the points between the two points that make the line thatdetermines the half-planes, H1 and H2. With this script, I had to use the arc on circle to define thearc for illustrative purposes. Therefore, segment AB is congruentto segment CD, since opposite sides of a parallelogram are equal. Since line l is parallel to line k and line a is parallel to line b,ABCD is a parallelogram, by definition. This construction is possible using the three Euclidean constructionrules. Since corresponding partsof congruent triangles are congruent, angle CBF is congruent to angle ABF.By the definition of angle bisector, ray BF is an angle bisector of angleCBF. By SSS, triangles BCF and BAF are similar. SegmentBF is congruent to BF, since it is a common side of the triangles BCF andBAF. Similarly, segment CF is congruent to segment AF. Since segment BC and AB are radii of circle B, then segment BC is congruentto segment AB. Here ray BF is the angle bisector of angle ABC. It is possible to construct the angle bisector using the three Euclideanconstruction rules as stated above. In the figure above, line m isa transversal and the corresponding angles both measure 90 degrees, sincethe lines were perpendicular to each other. This theorem directly followsfrom the Corresponding Angles Postulate that says, Suppose two coplanarlines are cut by a transversal and if two corresponding angles have thesame measure, then the lines are parallel. Therefore, we can say that line kis parallel to line j, based on this theorem. If two coplanar lines j and k are perpendicular to the same line,then they are parallel to each other. It is possible to construct the parallel to a given line and a pointnot on the line using the three Euclidean rules as stated above.īased on the past explanations of why the preceeding constructions work,we constructed line m perpendicular to line j and then line k perpendicularto line m. We are ableto construct the perpendicular line k, through A, to line j. We bgein with line j and the point A that lies on line j. See above for explicit reasons.Ĭase II: Construct the perpendicular to a given line and apoint on the line. The reasoning process is the same as above for the midpoint of a segmentand perpendicular bisector. Sinceangle AEC and angle BEC form a linear pair and the angles are congruentto one another, it follows that angle AEC and angle BEC measure 90 degrees.Therefore, by definition of perpendicular, line AB is perpendicular to lineEC.Ĭase I: Construct the perpendicular to a given line and a point not on the line. Therefore, by correspondingparts of congruent triangles, angle AEC is congruent to angle BEC. Therefore,by definition of perpendicular lines, line j is perpendicular to line k.įollowing on the same argument above for the midpoint, we know that triangleACE is congruent to triangle BCE (see above). We know angle EBC measures 60 degrees, ECBmeasures 30 degrees, thus, angle CEB must measure 90 degrees. Since line CE bisects angle ACB,angle ECB measures 30 degrees. Therefore, each angle measures60 degrees, angle A, angle B, and angle C. Also, since segments AC, BC and AB are congruentto each other, triangle ABC is equilateral. Segments AC, BC and AB are congruent to each other, since they are allradii of circle A and circle B. In the following picture, line j is perpendicular to line k. This case is also a special caseof Case II below.) It is possible to construct a perpendicular line using the three Euclideanrules above. Since corresponding parts ofcongruent triangles are congruent, segment AM is congruent to segment CM.Hence, by definition of midpoint, M is the midpoint of segment AC. Then, by SAS,triangle ABM is congruent to triangle CBM. Therefore, triangles ABD and CBD are congruent, by SSS. Segments AB, BC, CD, AD are all congruent, since they are all radii ofthe circles A and C whose radii are equal by construction. Point at Midpoint is possible to construct using only the Euclidean rulesabove. However, simply picking an arbitrarypoint to construct on an oject cannot be done. It could be done if you were considering the pointon object to be a point at intersection. It is impossible to construct a point on an object using the three Euclideanconstructions above. More precisely, for which of the items in the "Construction"menu can a script be written using only the items Which constructions in GSP can be done using only the three Euclideanconstruction rules? Assignment 14-Euclidean Constructions using Straightedge and CompassĮuclidean Constructions using Straightedgeand Compass
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