Question 468892
ABC is an right angled triangle,right angled at B.Internal bisector of angle B meet AC at D.AB=3cm,BC=4 cm, find AD.
<pre>
{{{drawing(1000/3,400,-1,4,-1,5,  

locate(0,0,A), locate(3,0,B), locate(3,4+.2,C),locate(9/7-.2,12/7+.2,D), 

locate(3.1,2,4), locate(1.5,0,3), locate(2.35,.3,"45°"),
triangle(0,0,3,0,3,4), line(3,0,9/7,12/7)  )}}} 

Looking at the big right triangle ABC, and the
fact that the tangent is {{{OPPOSITE/(ADJACENT)}}}

{{{tan("<A") = (BC)/(AB) = 4/3}}}

Use the inverse tangent on a calculator to find &#8736;A:

&#8736;A = 53.13010235°

We know that &#8736;ABD is 45° because BD bisects the right angle at B

We find &#8736;ADB by using the fact that
the sum of the interior angles of &#4416;ABD is 180°

&#8736;A + &#8736;ABD + &#8736;ADB = 180° 

&#8736;ADB = 180° - &#8736;A - &#8736;ABD

&#8736;ADB = 180° - 53.13010235° - 45° = 81.86989765°

So we use the law of sines:

{{{(AD)/sin("<ABD")=(AB)/sin("<ADB")}}}

Cross-multiplying:

{{{AD*sin("<ADB")=AB*sin("<ABD")}}}

Divide both sides by {{{sin("<ADB")}}}:
  
{{{AD=(AB*sin("<ABD"))/sin("<ADB")}}}

{{{AD=(3sin("45°"))/sin("81.86989765°")}}}

Work out the right side with a calculator:

AD = 2.142857143cm

Edwin</pre>