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<!-- Mirrored from www.mathsisfun.com/geometry/circle-theorems.html by HTTrack Website Copier/3.x [XR&CO'2014], Sat, 29 Oct 2022 00:42:56 GMT -->
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<article id="content" role="main">
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Circle Theorems</h1>
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<p class="center">Some interesting things about angles and circles</p>
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<h2>Inscribed Angle</h2>
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<p>First off, a definition:</p>
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<div class="def">
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<p class="center"><b>Inscribed Angle</b>: an angle made from points sitting on the circle's circumference.</p>
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</div>
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<p class="center"><img src="images/inscribed-angle.svg" alt="inscribed angle ABC" height="194" width="198"><br>
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<span class="larger">A and C are "end points"<br>
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B is the "apex point"</span></p>
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<p>Play with it here:</p>
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<div class="script" style="height: 260px;">
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images/circle-prop.js?mode=inscribe
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</div>
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<p class="center">When you move point "B", what happens to the angle?</p>
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<h2>Inscribed Angle Theorems</h2>
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<p>Keeping the end points fixed ...</p>
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<p class="center">... the angle <span class="large">a°</span> is <b>always the same</b>,<br>
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no matter where it is on the <b>same arc</b> between end points:</p>
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<p class="center"><img src="images/inscribed-angle-2.svg" alt="inscribed angle always a on arc" height="173" width="173"><br>
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(Called the <b>Angles Subtended by Same Arc Theorem</b>)</p>
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<p><b>And</b> an inscribed angle <span class="large">a°</span> is half of the central angle <span class="large">2a°</span></p>
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<p class="center"><img src="images/inscribed-angle-1.svg" alt="inscribed angle a on circumference, 2a at center" height="173" width="173"><br>
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(Called the <b>Angle at the Center Theorem</b>) </p>
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<p>
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Try it here (not always exact due to rounding):
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</p>
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<div class="script" style="height: 260px;">
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images/circle-prop.js?mode=inscribe2
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</div>
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<div class="example">
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<h3>Example: What is the size of Angle POQ? (O is circle's center)</h3>
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<p class="center"><img src="images/inscribed-angle-ex1.svg" alt="inscribed angle 62 on circumference" height="155" width="167"></p>
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<p>Angle POQ = 2 × Angle PRQ = 2 × 62° = <b>124°</b></p>
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</div>
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<div class="example">
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<h3>Example: What is the size of Angle CBX?</h3>
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<p style="float:left; margin: 0 20px 5px 0;"><img src="images/inscribed-angle-ex2.svg" alt="inscribed angle example" height="167" width="175"></p>
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<p>Angle ADB = 32° also equals Angle ACB.</p>
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<p>And Angle ACB also equals Angle XCB.</p>
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<p>So in triangle BXC we know Angle BXC = 85°, and Angle XCB = 32°</p>
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<p>Now use angles of a triangle add to 180° :</p>
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<div class="so">Angle CBX + Angle BXC + Angle XCB = 180°</div>
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<div class="so">Angle CBX + 85° + 32° = 180°</div>
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<div class="so">Angle CBX = 63°</div>
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</div>
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<h2>Angle in a Semicircle (Thales' Theorem)</h2>
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<p>An angle <b>inscribed</b> across a circle's diameter is always a right angle:</p>
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<p class="center"><img src="images/angle-semicircle-1.svg" alt="angle inscribed across diameter is 90 degrees" height="173" width="173"><br>
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<i>(The end points are either end of a circle's diameter,<br>
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the apex point can be anywhere on the circumference.)</i></p>
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<p>Play with it here:</p>
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<div class="script" style="height: 260px;">
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images/circle-prop.js?mode=thales
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</div>
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<td valign="top">
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<p>Why? Because:</p>
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<p>The inscribed angle <span class="large">90°</span> is half of the central angle <span class="large">180°</span></p>
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<p>(Using "Angle at the Center Theorem" above)</p></td>
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<td><img src="images/angle-semicircle-1a.gif" alt="angle semicircle 90 degrees and 180 at center" height="180" width="182"></td>
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</tr>
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</tbody></table>
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<p> </p>
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<h3>Another Good Reason Why It Works</h3>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/angle-semicircle-2.gif" alt="angle semicircle rectangle" height="180" width="184"></p>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/angle-semicircle-3.gif" alt="angle semicircle rectangle" height="182" width="183"></p>
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<p>We could also rotate the shape around 180° to make a rectangle!</p>
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<p>It <b>is</b> a rectangle, because all sides are parallel, and both diagonals are equal.</p>
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<p>And so its internal angles are all right angles (90°).</p>
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<div class="example">
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<h3>Example: What is the size of Angle BAC?</h3>
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<p class="center"><img src="images/inscribed-angle-ex3.svg" alt="inscribed angle example" height="162" width="183"></p>
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<p>The Angle in the Semicircle Theorem tells us that Angle ACB = 90°</p>
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<p>Now use angles of a triangle add to 180° to find Angle BAC:</p>
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<div class="so"> Angle BAC + 55° + 90° = 180°</div>
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<div class="so"> Angle BAC = 35°</div>
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</div>
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<p> </p>
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<p class="center"><img src="images/angle-semicircle-4.gif" alt="angle semicircle always 90 on circumference" height="98" width="183"><br>
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So there we go! No matter <b>where</b> that angle is<br>
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on the circumference, it is <b>always 90°</b></p>
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<h3>Finding a Circle's Center</h3>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/circle-theorems-diameters.svg" alt="finding as circles center" height="173" width="173"></p>
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<p>We can use this idea to find a circle's center:</p>
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<ul>
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<li>draw a right angle from anywhere on the circle's circumference, then draw the diameter where the two legs hit the circle</li>
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<li>do that again but for a different diameter</li>
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</ul>
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<p>Where the diameters cross is the center!</p>
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<p><br></p>
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<h3>Drawing a Circle From 2 Opposite Points</h3>
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<p>When we know two opposite points on a circle we can draw that circle.</p>
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<p>Put some pins or nails on those points and use a builder's square like this:</p>
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<p class="center"> <img src="images/circle-theorems-thales-draw.svg" alt="finding as circles center" height="218" width="236"></p>
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<h2>Cyclic Quadrilateral</h2>
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<table style="border: 0;">
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<tbody>
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<tr>
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<td valign="top">
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<p align="right">A "Cyclic" <a href="../quadrilaterals.html">Quadrilateral</a> has every vertex on a circle's circumference:</p></td>
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<td><img src="images/quadrilateral-cyclic-1.svg" alt="quadrilateral cyclic" height="173" width="173"></td>
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</tr>
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<tr>
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<td valign="top">
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<p id="firstHeading3">A Cyclic Quadrilateral's <b>opposite angles add to 180°</b>:</p>
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<div class="center">
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a + c = 180°<br>
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b + d = 180°
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</div></td>
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<td><img src="images/quadrilateral-cyclic-2.svg" alt="quadrilateral cyclic a and c add to 180" height="173" width="173"></td>
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</tr>
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</tbody></table>
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<div class="example">
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<h3>Example: What is the size of Angle WXY?</h3>
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<p class="center"><img src="images/inscribed-angle-ex4.svg" alt="inscribed angle example" height="154" width="165"></p>
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<p>Opposite angles of a cyclic quadrilateral add to 180°</p>
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<div class="so"> Angle WZY + Angle WXY = 180°</div>
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<div class="so"> 69° + Angle WXY = 180°</div>
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<div class="so"> Angle WXY = 111°</div>
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</div><p> </p>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/angle-tangent.svg" alt="90 degrees between radius and tangent" height="232" width="227"></p>
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<h2>Tangent Angle</h2>
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<p>A <a href="tangent-secant-lines.html">tangent line</a> just touches a circle at one point.</p>
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<p>It always forms a right angle with the circle's radius.</p>
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<div style="clear:both">
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</div>
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<p> </p>
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<div class="questions">1012, 3240, 1013, 3241, 1014, 3242, 1015, 3243, 1016, 3244</div>
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<div class="related">
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<a href="circle.html">Circle</a>
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<a href="index.html">Geometry Index</a>
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<div id="copyrt">Copyright © 2022 Rod Pierce</div>
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