{"id":138,"date":"2017-08-23T09:06:56","date_gmt":"2017-08-23T09:06:56","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/mathematics\/?page_id=138"},"modified":"2017-08-30T07:28:02","modified_gmt":"2017-08-30T07:28:02","slug":"straightedge-and-compass-constructions","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/mathematics\/straightedge-and-compass-constructions\/","title":{"rendered":"Straightedge and Compass Constructions"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/quadrilaterals\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/geometry-spatial-reasoning\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/skew-lines\">Next Lesson \u27a1<\/a><\/div>\n<p><!-- CONTENT BEGINS HERE --><\/p>\n<h1 id=\"title\">Straightedge and Compass Constructions<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson, you will study definitions for the following objects: complementary and supplementary angles, angle\u00a0bisectors, and perpendicular bisector of a line segment. You will also construct these objects using a straightedge\u00a0and compass.<\/p>\n<h4>Previously Covered:<\/h4>\n<ul>\n<li>If two lines are cut by a transversal and any pair of\u00a0alternate interior angles is congruent, then the lines are\u00a0parallel.<\/li>\n<\/ul>\n<section>\n<h3><strong>What are the theorems involving parallel and perpendicular lines?<\/strong><\/h3>\n<p>Existence and Uniqueness of Parallel Lines\u00a0Let L be any line and P be a point not on L. Then there is<br \/>\nonly one line containing P parallel to L.\u00a0There is also an analogue to the theorem above for\u00a0perpendicular lines.<\/p>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li>Existence and Uniqueness of Perpendicular Lines<\/li>\n<li>Let L be any line and P be a point not on L. Then there is\u00a0only one line containing P perpendicular to L.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Although we may have implicitly referred to supplementary and\u00a0complementary angles in previous sections, let us define them here\u00a0for completeness.<\/p>\n<p>If the sum of the measures of two angles is 180\u00b0, then the\u00a0angles are called <abbr title=\"any pair of angles such that the sum of their measures is 180 degrees\">supplementary<\/abbr>,\u00a0and each angle is called the <strong><em>supplement\u00a0<\/em><\/strong>of the other.<\/p>\n<p>If the sum of the measures of two angles is 90\u00b0, then the\u00a0angles are called <abbr title=\"any pair of angles such that the sum of their measures is 90 degrees\">complementary<\/abbr> and each angle is called the <strong><em>complement\u00a0<\/em><\/strong>of the other.<\/p>\n<h4><strong>How do we construct an angle supplement?<\/strong><\/h4>\n<p>In this section, we will be dealing with angles formed by <em><strong>rays<\/strong><\/em>,\u00a0which are parts of a line that start at a point and extend\u00a0infinitely in the other direction. Below,\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/><em>BAC\u00a0<\/em>is formed by the rays\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p2_html_161136c.gif\" width=\"27\" height=\"21\" name=\"graphics16\" align=\"absmiddle\" border=\"0\" \/>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p2_html_m7137029a.gif\" width=\"28\" height=\"23\" name=\"graphics17\" align=\"absmiddle\" border=\"0\" \/>\u00a0. We\u2019ll start off with the simplest construction\u2014\u00a0angle supplements.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20001.JPG\" alt=\"Angle formed by two rays\" width=\"92\" height=\"104\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>Suppose we are given an angle <em>BAC\u00a0<\/em>as above where m<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics5\" align=\"absmiddle\" border=\"0\" \/><em>BAC\u00a0<\/em>&lt; 180\u00b0. We can construct the supplement to\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/><em>BAC\u00a0<\/em>with the same vertex <em>A\u00a0<\/em>using a straightedge.<\/p>\n<p><strong>Step 1:<\/strong> Extend the line containing the ray\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p2_html_161136c.gif\" width=\"27\" height=\"21\" name=\"graphics18\" align=\"BOTTOM\" border=\"0\" \/>\u00a0in the opposite direction using your straightedge.<\/p>\n<p><strong>Step 2:<\/strong> Choose a point <em>D\u00a0<\/em>on this new ray.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20002.JPG\" alt=\" Supplementary Angle\" width=\"144\" height=\"104\" name=\"graphics7\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>The angle <em>DAC<\/em> that we have just constructed is the\u00a0supplementary angle to\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p2_html_5029702c.gif\" width=\"17\" height=\"16\" name=\"graphics19\" align=\"BOTTOM\" border=\"0\" \/><em>BAC\u00a0<\/em>because\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s8_p2_clip_image004.gif\" width=\"176\" height=\"11\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p>The <abbr title=\" a line segment that divides an angle into two congruent angles \">angle\u00a0bisector<\/abbr> of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics9\" align=\"ABSMIDDLE\" border=\"0\" \/><em>BAC\u00a0<\/em>is the ray\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p2_html_mb09fd5c.gif\" width=\"28\" height=\"21\" name=\"graphics20\" align=\"BOTTOM\" border=\"0\" \/>,\u00a0such that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/><em>BAD\u00a0<\/em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/cong.gif\" width=\"14\" height=\"13\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics12\" align=\"ABSMIDDLE\" border=\"0\" \/><em>DAC<\/em>.\u00a0In other words, an angle bisector divides an angle into two\u00a0congruent angles.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20003.JPG\" alt=\" Supplementary Angle\" width=\"106\" height=\"104\" name=\"graphics13\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<h3><strong>How do we construct an angle bisector of a given angle?<\/strong><\/h3>\n<p>Suppose we are given\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/><em>A<\/em>.<\/p>\n<p><strong>Step 1<\/strong>: Using\u00a0<em>A\u00a0<\/em>as a center we draw an arc of a circle intersecting the two sides\u00a0of the angle. Label these points <em>B\u00a0<\/em>and <em>C<\/em>.\u00a0The segments\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p3_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics17\" align=\"BOTTOM\" border=\"0\" \/>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p3_html_m76b936f1.gif\" width=\"28\" height=\"23\" name=\"graphics18\" align=\"BOTTOM\" border=\"0\" \/>\u00a0are congruent\u00a0(these are both radii of the circle).<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20004.JPG\" alt=\"Angle A with circle intersecting the two sides\" width=\"128\" height=\"125\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 2:<\/strong> Draw\u00a0the circle with center <em>B\u00a0<\/em>and radius <em>r <\/em>=\u00a0<em>BC<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20005.JPG\" alt=\"Angle A with two circles intersecting\" width=\"155\" height=\"128\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 3: <\/strong>Draw\u00a0the circle with center <em>C\u00a0<\/em>and radius <em>r\u00a0<\/em>= <em>BC<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20006.JPG\" alt=\"circle with center C and r=BC\" width=\"155\" height=\"149\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 4:<\/strong> These\u00a0two circles will intersect at exactly two points. Choose the point\u00a0on the other side of the segment <em>BC\u00a0<\/em>and label this point <em>P<\/em>.\u00a0Draw the ray originating at <em>A\u00a0<\/em>through <em>P<\/em>.\u00a0This ray is the angle bisector of \u00a0<em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>BAC<\/em>.<\/p>\n<p align=\"CENTER\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20007.JPG\" alt=\"Angle bisector of angle BAC\" width=\"155\" height=\"149\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/><\/strong><\/p>\n<div class=\"callout\">\n<h4>Important Tidbit<\/h4>\n<p>In steps 2 and 3, we could have used any radius greater than <img loading=\"lazy\" decoding=\"async\" class=\"notebox_text\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s8_p3_clip_image002.gif\" width=\"26\" height=\"36\" name=\"graphics9\" align=\"ABSMIDDLE\" border=\"0\" \/>. This ensures that the circles will intersect.<\/p>\n<\/div>\n<p>Let&#8217;s try an example. When constructing an angle on a given\u00a0side of a ray congruent to a given <em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/>A<\/em>,<br \/>\nwhat is the minimum number of arcs (or circles) necessary to\u00a0construct the angle?<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/MAth%20Mod%204.3%20Art%20008.JPG\" alt=\"Construction of an angle\" width=\"238\" height=\"104\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>The correct answer is 3. We need to draw the first arc on the\u00a0given angle <em>A\u00a0<\/em>to create a triangle <em>ABC\u00a0<\/em>with congruent sides <em>r\u00a0<\/em>= <em>AB <\/em>=\u00a0<em>AC\u00a0<\/em>and a third side <em>BC\u00a0<\/em>=<em> s<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/MAth%20Mod%204.3%20Art%20009.JPG\" alt=\" Angle Construction part 2\" width=\"128\" height=\"118\" name=\"graphics12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>We draw a circle of radius\u00a0<em>r <\/em>with\u00a0center <em>D\u00a0<\/em>that intersects the given ray at a point <em>E<\/em>.\u00a0This gives us one side of a triangle that will eventually be\u00a0congruent to <em>BAC<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20010.JPG\" alt=\" Construction of an angle Part 3\" width=\"347\" height=\"119\" name=\"graphics13\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>Next, we draw a circle of\u00a0radius <em>s\u00a0<\/em>and center <em>E<\/em>.\u00a0This intersects this first circle in two points, but we choose the\u00a0one on the given side. The ray drawn through this point (F)\u00a0creates the desired congruent triangle (by SSS), and thus the\u00a0corresponding angles\u00a0<em>A<\/em> and <em>D\u00a0<\/em>we have constructed are also congruent.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20011.JPG\" alt=\"Angle Construction final step\" width=\"364\" height=\"137\" name=\"graphics14\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<h3><strong>How do we construct a line parallel to a given line containing an external given point?<\/strong><\/h3>\n<p>Suppose we are given a line\u00a0<em>L\u00a0<\/em>and a point <em>P\u00a0<\/em>that is not on <em>L<\/em>. Let <em>Q\u00a0<\/em>and <em>R\u00a0<\/em>be any two points on <em>L<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20012.JPG\" alt=\" Line L with point P\" width=\"158\" height=\"115\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 1:<\/strong> To\u00a0construct a parallel line to point <em>P<\/em>,\u00a0first draw the line\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p4_html_m39707f5.gif\" width=\"27\" height=\"25\" name=\"graphics8\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20013.JPG\" alt=\"Line L and line PQ\" width=\"158\" height=\"148\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 2:<\/strong> We\u00a0can draw angle <em>QPS\u00a0<\/em>congruent to angle <em>PQR\u00a0<\/em>on the opposite side of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p4_html_m39707f5.gif\" width=\"27\" height=\"25\" name=\"graphics9\" align=\"absmiddle\" border=\"0\" \/>\u00a0from <em>R<\/em>. These two angles, <em>QPS\u00a0<\/em>and <em>PQR<\/em>, are alternate interior angles.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20014.JPG\" alt=\"Alternate interior angles\" width=\"171\" height=\"148\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>Because these angles are\u00a0congruent, the line\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p4_html_1d150818.gif\" width=\"25\" height=\"23\" name=\"graphics10\" align=\"absmiddle\" border=\"0\" \/>\u00a0is parallel to the line <em>L<\/em>.<strong><em><br \/>\n<\/em><\/strong><\/p>\n<h3><strong>How do we construct a perpendicular bisector?<\/strong><\/h3>\n<p>The <abbr title=\" the line perpendicular to the segment at its midpoint\">perpendicular\u00a0bisector<\/abbr> of a line segment is the line perpendicular\u00a0to the segment at its midpoint.<\/p>\n<p>Suppose we are given a line segment\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p5_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p><strong>Step 1<\/strong> : To\u00a0construct a perpendicular bisector, first draw a circle with\u00a0center <em>B\u00a0<\/em>and radius <em>r <\/em>=\u00a0<em>AB<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20015.JPG\" alt=\"Circle with radius AB\" width=\"162\" height=\"148\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 2:<\/strong> Draw\u00a0a circle with center at <em>A\u00a0<\/em>and radius <em>r\u00a0<\/em>= <em>AB<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20016.JPG\" alt=\"Circle with center B and radius AB\" width=\"226\" height=\"151\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 3:<\/strong> The\u00a0two circles intersect at two points on opposite sides of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p5_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/>.\u00a0Label these points <em>P\u00a0<\/em>and <em>Q<\/em>.\u00a0Draw\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p5_html_b6bda4d.gif\" width=\"27\" height=\"25\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20017.JPG\" alt=\"Two circles intersect at two points on opposite sides of AB\" width=\"226\" height=\"188\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><em>PQ <\/em>is\u00a0equidistant from <em>A\u00a0<\/em>and <em>B<\/em>,\u00a0and is, therefore, the perpendicular bisector of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p5_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics10\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<h3><strong>How do we construct a line perpendicular to a given line containing an external given point?<\/strong><\/h3>\n<p>The process of constructing\u00a0a line perpendicular to a given line containing an external point\u00a0is quite similar to the construction we have just completed.\u00a0Basically, all we need to do is find the right segment to bisect.\u00a0So let\u2019s just do this one step.<\/p>\n<p>Again suppose we are given\u00a0a line <em>L <\/em>and\u00a0a point <em>P\u00a0<\/em>that is not on <em>L<\/em>.\u00a0Draw a circle with center <em>P\u00a0<\/em>and with large enough radius so that it intersects <em>L\u00a0<\/em>in two points, <em>Q\u00a0<\/em>and <em>R\u00a0<\/em>. If we construct the perpendicular bisector of <em>QR<\/em>,\u00a0the line passes through <em>P,\u00a0<\/em>because <em>P\u00a0<\/em>is equidistant from <em>R\u00a0<\/em>and <em>S<\/em>.\u00a0So by the previous construction, the perpendicular bisector of <em>QR\u00a0<\/em>contains <em>P\u00a0<\/em>and is perpendicular to <em>L<\/em>.<\/p>\n<h4><strong>How do we construct an angle complement?<\/strong><\/h4>\n<p>If you recall, we gave the definition of an angle <abbr title=\"any pair of angles such that the sum of their measures is 90 degrees\">complement<\/abbr> at the very beginning of the section. This construction follows\u00a0from work we\u2019ve already practiced.<\/p>\n<p>The construction of an\u00a0angle complementary to a given\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/angle.gif\" width=\"17\" height=\"16\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/><em>A\u00a0<\/em>sharing the same vertex, <em>A,\u00a0<\/em>is a corollary of the construction of a perpendicular bisector.<\/p>\n<p>If we extend one ray of\u00a0angle <em>A\u00a0<\/em>to a line and create a segment <em>BC<\/em>,\u00a0such that <em>A\u00a0<\/em>is the midpoint of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p6_html_786a0686.gif\" width=\"27\" height=\"23\" name=\"graphics7\" align=\"BOTTOM\" border=\"0\" \/>,\u00a0then the perpendicular bisector of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p6_html_786a0686.gif\" width=\"27\" height=\"23\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/>\u00a0will give us a\u00a0ray, <em>AD, <\/em>which\u00a0creates the angle complement to angle <em>A<\/em>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20018.JPG\" alt=\"Angle Complement\" width=\"137\" height=\"158\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<h3><strong>How do we divide a given line segment into <em>n\u00a0<\/em>congruent subsegments?<\/strong><\/h3>\n<p>Suppose we are given a line\u00a0segment\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p7_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/>\u00a0that we wish to\u00a0subdivide into <em>n\u00a0<\/em>congruent subsegments. The figure below shows the case for <em>n\u00a0<\/em>= 5.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20019.JPG\" alt=\"Line segment divided into n=5 subsegments\" width=\"339\" height=\"162\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 1:<\/strong> Draw\u00a0any ray originating at the point <em>A<\/em>,\u00a0but not containing <em>B<\/em>.\u00a0Name this ray\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p7_html_21d19341.gif\" width=\"25\" height=\"21\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><strong>Step 2:<\/strong> Draw\u00a0an arc of a circle centered at <em>A\u00a0<\/em>with positive radius <em>r\u00a0<\/em>and intersecting <em>L\u00a0<\/em>at a point <em>P<sub>1<\/sub><\/em>.<\/p>\n<p><strong>Step 3:\u00a0<\/strong>Draw an arc of a circle centered at <em>P<sub>1\u00a0<\/sub><\/em>with the same radius <em>r\u00a0<\/em>and intersecting <em>L\u00a0<\/em>at a point <em>P<sub>2<\/sub><\/em>.<\/p>\n<p><strong>Step 4:<\/strong> Repeat\u00a0step 3 until you reach <em>P<sub>n\u00a0<\/sub><\/em>. At this point you\u2019ve drawn <em>n\u00a0<\/em>congruent segments\u00a0on <em>L:P<sub>1<\/sub>,\u00a0P<sub>1<\/sub>P<sub>2<\/sub>, P<sub>2<\/sub>P<sub>3<\/sub>, . . . P<sub>n\u00a0\u2013 1<\/sub>P<sub>n<\/sub><\/em>.<\/p>\n<p><strong>Step 5:<\/strong> Using\u00a0your straightedge, connect <em>P<sub>n\u00a0<\/sub><\/em>to <em>B<\/em>.<\/p>\n<p><strong>Step 6:<\/strong> Through\u00a0the other points <em>P<sub>1<\/sub>,\u00a0P<sub>2<\/sub><\/em>, . .\u00a0.<em>P<sub>n \u2013 1<\/sub><\/em>,\u00a0draw lines parallel to <em>P<sub>n\u00a0<\/sub><\/em><em>B,\u00a0<\/em>intersecting\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p7_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/>\u00a0at points <em>Q<\/em><sub>1<\/sub>,\u00a0<em>Q<\/em><sub>2<\/sub>,&#8230;<em>Q<sub>n\u20131<\/sub><\/em>,\u00a0respectively.<\/p>\n<p>Because the parallel lines\u00a0intercept congruent segments on <em>L<\/em>,\u00a0they intersect congruent segments on\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s8_p7_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics7\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/quadrilaterals\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/geometry-spatial-reasoning\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/skew-lines\">Next Lesson \u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Straightedge and Compass Constructions Objective In this lesson, you will study definitions for the following objects: complementary and supplementary angles, angle\u00a0bisectors, and perpendicular bisector of a line segment. You will also construct these objects using a straightedge\u00a0and compass. Previously Covered: If two lines are cut by a transversal and [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-138","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/138","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/comments?post=138"}],"version-history":[{"count":11,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/138\/revisions"}],"predecessor-version":[{"id":520,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/138\/revisions\/520"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/media?parent=138"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}