Question 1035576
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In a triangle ABC ,AB=12cms , BC=8cms, CA=10cms. An incircle is drawn inside the triangle touching the sides AB, BC and CA 
at the points D, E and F respectively. Find AF, BD, and CE.
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<pre><TABLE>
  <TR>
  <TD>
I have found an appropriate figure from my archive &nbsp;(see on the right).
Although it is not in the scale, it is not so important for the solution. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

The names a, b and c are just employed for the side lengths:
|AB| = c = 12 cm, |BC| = a = 8 cm, and |AC| = b = 10 cm.

So we introduce d = |AD|= |AF|, e = |BE| = |BD| and f = |CF| = |CE|.

I hope you are familiar with the fact that tangent lines to a circle drawn 
from the outside point, are congruent, and therefore doubled equalities 
in the line above do not arose questions.

We have this system of equations

|AD| + |BD| = |AB|,                                 d + e = a,    (1)
|BE| + |CE| = |BC|,       or, which is the same,    e + f = b.    (2)
|AF| + |CF| = |AC|                                  d + f = c.    (3)

By adding equations (1), (2) and (3), you get

2(d + e + f) = a + b + c,   or    d + e + f = {{{(a + b + c)/2}}}.     (4)
 </TD>
  <TD>
{{{drawing( 300, 300,  0, 6, 0, 6, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

            locate(1.0, 1.0, A),
            locate(5.0, 1.0, B),
            locate(4.0, 5.4, C),

            locate(2.6, 1.0, c),

            locate(4.6, 3.2, a),

            locate(2.05, 3.05, b),

      green(line(1.0, 1.0, 3.4, 2.22)),
      green(line(4.0, 5.0, 3.4, 2.22)),
      green(line(5.0, 1.0, 3.4, 2.22)),

            circle(3.35, 2.21, 0.06, 0.06),
            locate(3.20, 2.7, P),

        red(circle(3.4, 2.22, 1.2, 1.2)),

        red(line (3.38, 2.22, 3.38, 1.0)),
            locate(3.45, 1.9, r),

        red(line (3.35, 2.22, 2.46, 2.95)),
            locate(2.7, 2.67, r),

        red(line (3.35, 2.20, 4.5, 2.6)),
            locate(4.1, 2.5, r),

            locate(3.3, 1.0, D),
            locate(4.7, 2.8, E),
            locate(2.3, 3.2, F)
)}}}
 </TD>
 </TR>
</TABLE>Now distract equation (1) from (4).         You will get  f = {{{(a + b + c)/2 - a}}} = {{{(-a+b+c)/2}}}.    (5)

Next, distract equation (2) from (4).       You will get  d = {{{(a + b + c)/2 - b}}} = {{{(a-b+c)/2}}}.     (6)

Similarly, distract equation (3) from (4).  You will get  e = {{{(a + b + c)/2 - c}}} = {{{(a+b-c)/2}}}.     (7)

Now substitute (plug in) the given numerical data into equations (5), (6) and (7). You will get

f = {{{(-8+10+12)/2}}} = 7 cm,  d = {{{(8-10+12)/2}}} = 5 cm, e = {{{(8+10-12)/2}}} = 3 cm.

<U>Answer</U>.  AF = d = 5 cm,  BD = e = 3 cm, and CE = f = 7 cm.
</pre>

The problem is solved.