document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210139>Question 1210139</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the diagram, \overline{AD} is an altitude, \overline{BE} is a median, and \overline{AD}, \overline{BE}, and \overline{CF} are concurrent. If AB = 11, AC = 12, and AD = 6, what is AF?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851358 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851358\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $\triangle ABC$ be a triangle. Let $AD$ be an altitude, $BE$ be a median, and $CF$ be a cevian. We are given that $AD$, $BE$, and $CF$ are concurrent. Let $AB = 11$, $AC = 12$, and $AD = 6$. We want to find $AF$.\r<BR>\n" );
document.write( "\n" );
document.write( "Since $AD$, $BE$, and $CF$ are concurrent, we can apply Ceva's Theorem. Ceva's Theorem states that for a triangle $ABC$ with cevians $AD$, $BE$, and $CF$ concurrent, we have<BR>\n" );
document.write( "$$ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1 $$\r<BR>\n" );
document.write( "\n" );
document.write( "Since $BE$ is a median, $CE = EA$. Thus, $\frac{CE}{EA} = 1$.<BR>\n" );
document.write( "The equation becomes<BR>\n" );
document.write( "$$ \frac{AF}{FB} \cdot \frac{BD}{DC} = 1 $$<BR>\n" );
document.write( "$$ \frac{AF}{FB} = \frac{DC}{BD} $$\r<BR>\n" );
document.write( "\n" );
document.write( "We are given $AB = 11$, $AC = 12$, and $AD = 6$.<BR>\n" );
document.write( "In $\triangle ABD$, we have $BD^2 = AB^2 - AD^2 = 11^2 - 6^2 = 121 - 36 = 85$. Thus, $BD = \sqrt{85}$.<BR>\n" );
document.write( "In $\triangle ACD$, we have $CD^2 = AC^2 - AD^2 = 12^2 - 6^2 = 144 - 36 = 108$. Thus, $CD = \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}$.\r<BR>\n" );
document.write( "\n" );
document.write( "Now we have<BR>\n" );
document.write( "$$ \frac{AF}{FB} = \frac{6\sqrt{3}}{\sqrt{85}} $$<BR>\n" );
document.write( "Let $AF = x$. Then $FB = AB - AF = 11 - x$.<BR>\n" );
document.write( "$$ \frac{x}{11-x} = \frac{6\sqrt{3}}{\sqrt{85}} $$<BR>\n" );
document.write( "$$ x\sqrt{85} = 6\sqrt{3}(11-x) $$<BR>\n" );
document.write( "$$ x\sqrt{85} = 66\sqrt{3} - 6x\sqrt{3} $$<BR>\n" );
document.write( "$$ x(\sqrt{85} + 6\sqrt{3}) = 66\sqrt{3} $$<BR>\n" );
document.write( "$$ x = \frac{66\sqrt{3}}{\sqrt{85} + 6\sqrt{3}} $$<BR>\n" );
document.write( "Multiply the numerator and denominator by $\sqrt{85} - 6\sqrt{3}$:<BR>\n" );
document.write( "$$ x = \frac{66\sqrt{3}(\sqrt{85} - 6\sqrt{3})}{85 - 36 \cdot 3} = \frac{66\sqrt{3}(\sqrt{85} - 6\sqrt{3})}{85 - 108} = \frac{66\sqrt{3}(\sqrt{85} - 6\sqrt{3})}{-23} $$<BR>\n" );
document.write( "$$ x = \frac{66\sqrt{3}(6\sqrt{3} - \sqrt{85})}{23} = \frac{66(18 - \sqrt{255})}{23} $$<BR>\n" );
document.write( "$$ x = \frac{1188 - 66\sqrt{255}}{23} $$\r<BR>\n" );
document.write( "\n" );
document.write( "However, we can use Menelaus' Theorem on $\triangle ABD$ and line $FC$:<BR>\n" );
document.write( "$$ \frac{AF}{FB} \cdot \frac{BC}{CD} \cdot \frac{DG}{GA} = 1 $$<BR>\n" );
document.write( "We can also use Menelaus' Theorem on $\triangle ADC$ and line $FB$:<BR>\n" );
document.write( "$$ \frac{AF}{FC} = \frac{x}{CF} $$\r<BR>\n" );
document.write( "\n" );
document.write( "By Ceva's Theorem, $\frac{AF}{FB} = \frac{DC}{BD}$.<BR>\n" );
document.write( "$BD = \sqrt{11^2 - 6^2} = \sqrt{121-36} = \sqrt{85}$<BR>\n" );
document.write( "$CD = \sqrt{12^2 - 6^2} = \sqrt{144-36} = \sqrt{108} = 6\sqrt{3}$<BR>\n" );
document.write( "$\frac{AF}{11-AF} = \frac{6\sqrt{3}}{\sqrt{85}}$<BR>\n" );
document.write( "$AF\sqrt{85} = 66\sqrt{3} - 6\sqrt{3}AF$<BR>\n" );
document.write( "$AF(\sqrt{85} + 6\sqrt{3}) = 66\sqrt{3}$<BR>\n" );
document.write( "$AF = \frac{66\sqrt{3}}{\sqrt{85} + 6\sqrt{3}} \approx \frac{114.315}{9.2195 + 10.3923} = \frac{114.315}{19.6118} \approx 5.829$\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{6}$<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );