Find the area of ​the triangular region ABC

For reference;Calculate the area of ​​the triangular region ABC, if : $CK = a$ ($T$ is tangency point)9Answer:$a^2$}

My progress:

I'm in difficulties .. I don't see many alternatives $S_{ABC} = \frac{AB.BC}{2}\\ S_{ABC} = r(\frac{AB+BC+AC}{2}\\ s_{ABC} = AT.TC\\ TS^2+(2r)^2 = (2R)^2\implies TS=\sqrt{4R^2-4r^2}=2\sqrt{R^2-r^2}\\ $

...???

Solution 1:

In the usual notation, the incircle touches the sides of a triangle at $s-a,s-b,s-c$ from the vertices $A,B,C$ respectively.

$B$-excircle touches $AC$ at $(s-c)$ from $A$ and $(s-a)$ from $C$. So if $D$ were to be tangency point of $B$-excircle, then $CD=AF$ and $CF=AD$.

$B$-excircle is just a magnified version of incircle (or the two circles are homothetic wrt $B$). Every point on its circumference is collinear with corresponding point on the incircle and the vertex $B$. In particular, its tangency point on $AC$, which is its uppermost point, is in line with the uppermost point of incircle and $B$. From the collinearity of $B,E,D$, it turns out that $D$ is indeed the touch-point of the excircle.

Set $s-a=x$ and $s-c=y$. Lets observe :

• $CK^2=CD \times CF=xy$
• $x-y=(s-a)-(s-c)=c-a$

Now we are ready for the solution. $$(x+y)^2=(x-y)^2+4xy $$ $$\Rightarrow b^2 = (a-c)^2+4CK^2$$ $$\Rightarrow b^2=a^2+c^2-2ac+4CK^2$$ Since $b^2=a^2+c^2$ and $S_{ABC}=ac/2$, it follows that $$0=-4S_{ABC}+4CK^2$$ $$\therefore \boxed{S_{ABC}=CK^2}$$