Question 1087582
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triangle(0,0,12,0,12,4), red(arc(0,0,6,-6,0,19),arc(0,0,5.5,-5.5,19,37)),
locate(12.1,2.4,4),locate(6,0,12),locate(12.1,4.4,D),
locate(0,0,A),locate(12,0,B),locate(12,9.7,C),
triangle(0,0,12,0,12,9)  )}}}
<pre><b>
{{{tan("<BAD")}}}{{{""=""}}}{{{BD/(AB)}}}{{{""=""}}}{{{4/12}}}{{{""=""}}}{{{1/3}}}

Since AD is the angle bisector of &#8736;BAC, &#8736;BAC = 2&#8736;BAD.

Use formula {{{tan(2*theta)}}}{{{""=""}}}{{{2tan^""(theta)/(1-tan^2(theta))}}}

{{{tan("<BAC")}}}{{{""=""}}}{{{tan(2*"<BAD")}}}{{{""=""}}}{{{2tan^""("<BAD")/(1-tan^2("<BAD"))}}}{{{""=""}}}{{{2(1/3)/(1+(1/3)^2)}}}{{{""=""}}}{{{(2/3)/(1-1/9)}}}{{{""=""}}}{{{(2/3)/(8/9)}}}{{{""=""}}}{{{(2/3)(9/8)}}}{{{""=""}}}{{{18/24}}}{{{""=""}}}{{{3/4}}}

And since also {{{tan("<BAC")}}}{{{""=""}}}{{{OPPOSITE/ADJACENT=BC/(AB)}}}{{{""=""}}}{{{BC/12}}}

{{{BC/12}}}{{{""=""}}}{{{3/4}}}

{{{4*BC}}}{{{""=""}}}{{{36}}}

{{{BC}}}{{{""=""}}}{{{9}}}, (which means, incidentally that DC = 5)

By the Pythagorean theorem:

{{{AC^2}}}{{{""=""}}}{{{AB^2+BC^2}}}

{{{AC^2}}}{{{""=""}}}{{{12^2+9^2}}}

{{{AC^2}}}{{{""=""}}}{{{144+81}}}

{{{AC^2}}}{{{""=""}}}{{{225}}}

{{{AC}}}{{{""=""}}}{{{sqrt(225)}}}

{{{AC}}}{{{""=""}}}{{{15}}}

Edwin</pre></b>