<s>[INST] Finding $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}$ Suppose circle $I$ is $\triangle ABC$’s incircle and $D, E,$ and $F$ are the tangent points of $BC, CA, AB$ and circle $I.$ Given that $DI$ intersects $EF$ at $K, BC = 10, AC = 8,$ and $AB = 7$ find $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}.$

 [/INST] 
We use the identity $A = r \cdot s, $ where $s$ is sub-perimeter, $r$ is the inradius and $A$ is the area of the triangle.
So area of $\displaystyle \triangle ABC  = \frac{10+8+7}{2} \cdot r = \frac{25r}{2}$
Area of $\triangle BIC = \frac{1}{2} BC \cdot DI = 5r$
Area of $\triangle BIK = \frac{1}{2} BD \cdot KI$, Area of $\triangle CIK = \frac{1}{2} CD \cdot KI$
So all we are left with is to find $KI$. For that, we draw internal angle bisector of $\angle A$ which meets $BC$ at $M$.
As $AF = AE, FI = IE, \angle AFI = \angle AEI = 90^0$, therefore $AI$ is perpendicular bisector of $FE$.
So $\angle IFK = \frac{\angle A}{2}$. Also, $\angle FIK = 180^0 - \angle FID = \angle B$.
That leads to $\triangle FIK \sim \triangle ABM$ (by A-A-A)
$\frac{KI}{FI} = \frac{BM}{AB} = \frac{14/3}{AB} \implies KI = \frac{2r}{3}$
(to calculate $BM$, we used the fact that $AM$ is the angle bisector and hence $BM:MC = AB:AC = 7:8$ and $BM + MC = 10$)
Getting to concluding steps, area of $\triangle BKC = 5r + \frac{1}{2} (BD+CD) \frac{2r}{3} = \frac{25r}{3}$
$\therefore \frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{2}{3}$.
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