https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathwizard112358&feedformat=atom AoPS Wiki - User contributions [en] 2021-10-27T23:42:25Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Stewart%27s_Theorem&diff=103845 Stewart's Theorem 2019-02-26T04:26:15Z <p>Mathwizard112358: /* Statement */</p> <hr /> <div>== Statement ==<br /> Given a [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; with sides of length &lt;math&gt;a, b, c&lt;/math&gt; opposite [[vertex | vertices]] are &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, respectively. If [[cevian]] &lt;math&gt;AD&lt;/math&gt; is drawn so that &lt;math&gt;BD = m&lt;/math&gt;, &lt;math&gt;DC = n&lt;/math&gt; and &lt;math&gt;AD = d&lt;/math&gt;, we have that &lt;math&gt;b^2m + c^2n = amn + d^2a&lt;/math&gt;. (This is also often written &lt;math&gt;man + dad = bmb + cnc&lt;/math&gt;, a form which invites mnemonic memorization, i.e. &quot;A man and his dad put a bomb in the sink.&quot; When you're practicing to memorize this formula, never practice it in the library/airport or any other public place where other people can hear you.)<br /> <br /> &lt;center&gt;[[Image:Stewart's_theorem.png]]&lt;/center&gt;<br /> <br /> == Proof ==<br /> Applying the [[Law of Cosines]] in triangle &lt;math&gt;\triangle ABD&lt;/math&gt; at [[angle]] &lt;math&gt;\angle ADB&lt;/math&gt; and in triangle &lt;math&gt;\triangle ACD&lt;/math&gt; at angle &lt;math&gt;\angle CDA&lt;/math&gt;, we get the equations<br /> *&lt;math&gt; n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} &lt;/math&gt;<br /> *&lt;math&gt; m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2} &lt;/math&gt;<br /> <br /> Because angles &lt;math&gt;\angle ADB&lt;/math&gt; and &lt;math&gt;\angle CDA&lt;/math&gt; are [[supplementary]], &lt;math&gt;m\angle ADB = 180^\circ - m\angle CDA&lt;/math&gt;. We can therefore solve both equations for the cosine term. Using the [[trigonometric identity]] &lt;math&gt;\cos{\theta} = -\cos{(180^\circ - \theta)}&lt;/math&gt; gives us<br /> *&lt;math&gt; \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}&lt;/math&gt;<br /> *&lt;math&gt; \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}&lt;/math&gt;<br /> <br /> Setting the two left-hand sides equal and clearing [[denominator]]s, we arrive at the equation: &lt;math&gt; c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n &lt;/math&gt;.<br /> However, &lt;math&gt;m+n = a&lt;/math&gt; so &lt;math&gt;m^2n + n^2m = (m + n)mn&lt;/math&gt; and we can rewrite this as &lt;math&gt;man + dad= bmb + cnc&lt;/math&gt; (A man and his dad put a bomb in the sink). When you're practicing to memorize this formula, never practice it in the library or any other public place where other people can hear you.<br /> <br /> == See also == <br /> * [[Menelaus' Theorem]]<br /> * [[Ceva's Theorem]]<br /> * [[Geometry]]<br /> * [[Angle Bisector Theorem]]<br /> <br /> [[Category:Geometry]]<br /> <br /> [[Category:Theorems]]</div> Mathwizard112358