The angle bisectors
of the triangle meet at the center of the inscribed circle of radius

*r*. If we let \(2\alpha=A\), \(2\beta = B\), and \(2\gamma=C\), we have \(\alpha+\beta+\gamma=\frac{\pi}{2}\).
Let

Thus if we name the semiperimeter

*x*be the distance from the vertex at*A*to points of tangency,*y*the distance from*B*, and*z*the distance from*C*. Then then lengths of the triangle sides opposite*A*,*B*, and*C*are respectively \(a=y+z\), \(b=x+z\), and \(c=x+y\).Thus if we name the semiperimeter

*s*, then \(s=x+y+z\), \(x=s-a\), \(y=s-b\), and \(z=s-c\).
\(\tan \alpha =\frac{r}{x}\), \(\tan \beta =\frac{r}{y}\), and \(\tan \gamma =\frac{r}{z}\). Because \(\gamma\) and \( (\alpha+\beta )\) are complementary angles, we
obtain

\[ \tan\left( \frac{\pi}{2} - (\alpha+\beta) \right) = \frac{r}{z} \]

\[ \tan\left( \alpha+\beta \right) = \frac{z}{r} \]

\[ \frac{ \tan \alpha+\tan\beta}{1-\tan\alpha \tan\beta} = \frac{z}{r} \]

\[r \left(\tan \alpha+\tan\beta \right)= z (1-\tan\alpha\tan\beta) \]

\[r \left( \frac{r}{x}+\frac{r}{y} \right) = z \left( 1 - \frac{r}{x}\frac{r}{y} \right) \]

\[ r^2 y + r^2 z = xyz - r^2 z \]

\[ r^2 ( x+y+z) = xyz \]

\[ r^2 s = xyz \]

The radii at the
points of tangency and the angle bisectors form 3 pairs of congruent
triangles. The area of \(\Delta ABC\) is \(xr+yr+zr= r(x+y+z)\), so area \(=rs\), and \( (\text{area})^2=r^2s^2\). Using results we have above, we
obtain

\[ (\text{area})^2 = s\cdot xyz = s(s-a)(s-b)(s-c)\]

so the area is \(\sqrt{s(s-a)(s-b)(s-c)}\).

\[ (\text{area})^2 = s\cdot xyz = s(s-a)(s-b)(s-c)\]

so the area is \(\sqrt{s(s-a)(s-b)(s-c)}\).

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