Task A-3.2.

Let $ABC$ be a triangle in which $|AC| > |BC|$. Express the area of the triangle determined by side $\overline{AB}$, the perpendicular bisector of side $\overline{AB}$, and the angle bisector of $\angle ACB$ in terms of the lengths of the sides of triangle $ABC$.
## Solution.

Let the lengths of the sides and the measures of the angles of triangle $ABC$ be denoted by $a, b, c, \alpha, \beta, \gamma$ respectively.

Let point $P$ be the midpoint of side $\overline{AB}$, point $D$ the intersection of the given angle bisectors, and point $E$ the intersection of the angle bisector of $\angle ACB$ and side $\overline{AB}$. We need to express the area of triangle $DEP$ in terms of $a, b,$ and $c$. Since $DEP$ is a right triangle with legs $\overline{PD}$ and $\overline{PE}$, we have

$$
P(DEP) = \frac{|PD| \cdot |PE|}{2}
$$

By the Angle Bisector Theorem, $\frac{a}{b} = \frac{|BE|}{|AE|}$, and since $|AE| + |BE| = c$, we get $|BE| = \frac{ca}{a+b}$.

Since $P$ is the midpoint of side $\overline{AB}$, it follows that

$$
|PE| = |PB| - |EB| = \frac{c}{2} - \frac{ca}{a+b} = \frac{c(b-a)}{2(a+b)}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_a7570febbe914cc2404eg-16.jpg?height=759&width=744&top_left_y=1338&top_left_x=536)

Notice that both the angle bisector of $\angle ACB$ and the perpendicular bisector of side $\overline{AB}$ bisect the arc $\widehat{AB}$ of the circumcircle of triangle $ABC$, which implies that point $D$ is the midpoint of that arc, i.e., $D$ lies on the circumcircle of triangle $ABC$. By the Inscribed Angle Theorem applied to angles subtended by chord $\overline{DB}$, we have

$$
\angle DAP = \angle DAB = \angle DCB = \frac{\gamma}{2}.
$$

From the right triangle $PAD$, we conclude that

$$
|PD| = |AP| \tan \frac{\gamma}{2} = \frac{c}{2} \cdot \tan \frac{\gamma}{2}
$$

It remains to express $\tan \frac{\gamma}{2}$ in terms of $a, b, c$.

Let $s = \frac{a+b+c}{2}$ be the semiperimeter, $I$ the incenter of triangle $ABC$, and $J$ the projection of point $I$ onto side $\overline{BC}$. Then $|CJ| = \frac{a+b-c}{2} = s-c$, and from triangle $CIJ$ we conclude that $\tan \frac{\gamma}{2} = \frac{r}{s-c}$.

By equating the expressions for the area of the triangle

$$
\sqrt{s(s-a)(s-b)(s-c)} = P(ABC) = r \cdot s
$$

we get

$$
r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}
$$

so

$$
\tan \frac{\gamma}{2} = \frac{r}{s-c} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}
$$

Alternatively,

$$
\tan \frac{\gamma}{2} = \sqrt{\frac{1 - \cos \gamma}{1 + \cos \gamma}} = \sqrt{\frac{1 - \frac{a^2 + b^2 - c^2}{2ab}}{1 + \frac{a^2 + b^2 - c^2}{2ab}}} = \sqrt{\frac{2ab - a^2 - b^2 + c^2}{2ab + a^2 + b^2 - c^2}} = \sqrt{\frac{c^2 - (a-b)^2}{(a+b)^2 - c^2}}
$$

From (1), (2), (3), and (4), we get

$$
\begin{aligned}
P(DEP) & = \frac{1}{2} \cdot |PE| \cdot |PD| \\
& = \frac{1}{2} \cdot \frac{c(b-a)}{2(a+b)} \cdot \frac{c}{2} \cdot \tan \frac{\gamma}{2} \\
& = \frac{1}{2} \cdot \frac{c(b-a)}{2(a+b)} \cdot \frac{c}{2} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \\
& = \frac{c^2(b-a)}{8(a+b)} \sqrt{\frac{(b+c-a)(a+c-b)}{(a+b+c)(a+b-c)}}
\end{aligned}
$$