Radius of an e-circle in terms of triangle's sides and area. Following is the derivation given in my textbook:

I can't figure out that how do the areas of triangles $ABI_1$ and $ACI_1$ can be given by $\frac{1}{2} cr_1$ and $\frac{1}{2} br_1$ when $r_1$ is the radius of e-circle and $b,c$ are sides of triangle $ABC$ opposite to the vertices $B$ and $C$ respectively. 
$$Area = base \times height$$
So areas of triangles $ABI_1$ and $ACI_1$ should be given by $\frac{1}{2}.(\overline{AI_1}).(\overline{BP_1})$ and $\frac{1}{2}.(\overline{AI_1}).(\overline{CP_1})$ respectively. And of course (from the figure) it is clear that $$\overline{AI_1} \neq b$$ $$\overline{AI_1} \neq c$$ $$\overline{CP_1} \neq r_1$$ $$\overline{BP_1} \neq r_1$$
 A: Focus on the area of triangle $AF_1I_1$:
$$\text{area of }AF_1I_1 = \text{area of }ACI_1 + \text{area of }CF_1I_1$$
Note also that, since the circle is tangent to the lines in $F_1$ and $P_1$, those are right angles. Therefore:
$$\text{area of }ACI_1 = \frac12\overline{AF_1} r_1 - \frac12\overline{CF_1} r_1 = \frac12 \left(\overline{AF_1} - \overline{CF_1}\right)r_1 = \frac12 b r_1$$
A: Let the excircle at side AB touch at side AC extended at G, and let this excircle's radius be $ r_{c}$ and its center be $ {\displaystyle J_{c}}$.
Then $ {\displaystyle J_{c}G}$ is an altitude of $ {\displaystyle \triangle ACJ_{c}}$, so $ {\displaystyle \triangle ACJ_{c}}$ has area $ {\tfrac {1}{2}}br_{c}$. By a similar argument, ${\displaystyle \triangle BCJ_{c}}$ has area ${\tfrac {1}{2}}ar_{c}$ and $ {\displaystyle \triangle ABJ_{c}}$ has area $ {\tfrac {1}{2}}cr_{c}$. Thus the area $ \Delta$  of triangle ${\displaystyle \triangle ABC}$ is
$$ \Delta ={\frac {1}{2}}(a+b-c)r_{c}=(s-c)r_{c}.$$
So, by symmetry, denoting $r$ as the radius of the incircle,
$${\displaystyle \Delta =sr=(s-a)r_{a}=(s-b)r_{b}=(s-c)r_{c}}.$$
By the Law of Cosines, we have
$$\cos A={\frac {b^{2}+c^{2}-a^{2}}{2bc}}.$$
Combining this with the identity $\sin ^{2}A+\cos ^{2}A=1$, we have
$ \sin A={\frac {\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}}}{2bc}}$
But $\Delta ={\tfrac {1}{2}}bc\sin A$, and so
\begin{aligned}\Delta &={\frac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}}}\\&={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\&={\sqrt {s(s-a)(s-b)(s-c)}},\end{aligned}
which is Heron's formula.
Combining this with $sr=\Delta $, we have
$$ r^{2}={\frac {\Delta ^{2}}{s^{2}}}={\frac {(s-a)(s-b)(s-c)}{s}}.$$
Similarly, $(s-a)r_{a}=\Delta$  gives
$$r_{a}^{2}={\frac {s(s-b)(s-c)}{s-a}}$$
and
$$r_{a}={\sqrt {\frac {s(s-b)(s-c)}{s-a}}}.$$[24]
