https://blog.dlow.me/tags/ceva-theorem/Recent content in Ceva-Theorem on dlow's blogHugoen-usThu, 16 Jul 2020 09:00:15 +0100https://blog.dlow.me/maths/geometry-1/Thu, 16 Jul 2020 09:00:15 +0100https://blog.dlow.me/maths/geometry-1/<p>Found this while browsing and thought it was interesting. I&rsquo;m still deciding whether this should live on this blog. Don&rsquo;t permalink to this yet!</p> <p><figure> <a href="./20200716090210876_31970.png" rel="light"> <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" alt="" data-src="./20200716090210876_31970.png" class="lazyload" /> <noscript> <img src="./20200716090210876_31970.png" alt="" /> </noscript> </a> <figcaption> </figcaption> </figure> </p> <p><a href="https://gogeometry.blogspot.com/2008/05/geometry-problem-105.html">Source</a></p> <p>Technically there are two possible answers for x, but if you restrict ABC to be acute, then there is only one solution.</p> <p>Posed differently:</p> <ul> <li>Given an angle A and an internal angle bisector of A</li> <li>Let $A = 2 \beta $</li> <li>Choose a point D on the bisector</li> <li>Choose B on one leg of the angle such that $DBA = \alpha$</li> <li>Find the loci of C on the other leg such that $DCB = 90^\circ -\alpha - \beta$</li> </ul> <hr> <h2 id="solutions">Solutions: <a class="anchor" href="#solutions"><span></span></a></h2><p>Stolen from the comments from the original source: Using Ceva&rsquo;s theorem: We can simplify it to become $$ \frac{\sin(\alpha)}{\sin(\pi/2-\beta-x)} \cdot \frac{\sin(\pi/2 - \alpha - \beta)}{\sin(\beta)} = 1 $$</p>