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<title>Angle of Intersecting Secants Theorem</title>
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<h1 class="center">Angle of Intersecting Secants</h1>
<p>This is the idea (a,b and c are angles):</p>
<p class="center"><img src="images/circle-intersect-angle.svg" alt="Angle of Intersecting Secants" ></p>
<p>And here it is with some actual values:</p>
<p class="center"><img src="images/circle-intersect-angle-ex.svg" alt="Angle of Intersecting Secants example" ></p>
<p>In words: <i>the angle made by two <a href="tangent-secant-lines.html">secants</a> (a line that cuts a circle at two points) that <b>intersect outside</b> the circle is half of the furthest arc minus the nearest arc.</i></p>
<p>&nbsp;</p>
<p class="center larger">Why not try drawing one yourself, measure it using a protractor, <br>
and see what you get?</p>
<p>&nbsp;</p>
<p>It also works when either line is a <a href="tangent-secant-lines.html">tangent</a> (a line that just touches a circle at one point). Here we see the &quot;both are tangents&quot; case:</p>
<p class="center"><img src="images/circle-intersect-angle-ex2.svg" alt="Angle of Intersecting Secants example" ></p>
<p>That's it! You know it now.</p>
<p>&nbsp;</p>
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<h3>But How Come?</h3>
<p>Is this magic?</p>
<p>Well, we can prove it if you want:</p>
<p>AC and BD are two secants that intersect at the point P outside the circle. What is the relationship between the angle CPD and the arcs AB and CD?</p>
<p class="center"><img src="images/lines-circle-sec-sec-detail.svg" alt="Two lines intersecting a circle where vertex is inside the circle "></p>
<p>We start by saying that the angle subtended by arc CD at O is <b>2&theta;</b> and the arc subtended by arc AB at O is <b>2&Phi; </b></p>
<p>By the <a href="circle-theorems.html">Angle at the Center Theorem</a>: </p>
<p class="so">&ang;DAC = &ang;DBC = &theta; and &ang;ADB = &ang;ACB = &Phi; </p>
<p>And PAC is 180&deg;, so:</p>
<p class="so">&ang;DAP = 180&deg; &minus; &theta; </p>
<p>Now use <a href="../proof180deg.html">angles of a triangle add to 180&deg;</a> in triangle APD: </p>
<p class="so">&ang;CPD = 180&deg; &minus; (&ang;DAP + &ang;ADP) </p>
<p class="so">&ang;CPD = 180&deg; &minus; (180&deg; &minus; &theta; + &Phi;) = &theta; &minus; &Phi;</p>
<p class="so">&ang;CPD = &theta; &minus; &Phi;</p>
<p class="so">&ang;CPD = &frac12;(2&theta; &minus; 2&Phi;) </p>
<p>Done! </p>
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