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 $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}.$

 A: Since you let me have the pleasure of solving the problem (by not sharing your solution or approach), it is only fair that I reciprocate and let you enjoy the exercise of at least some parts of the proof. Here we go:
Lemma 1 - If in $\triangle ABC$ shown below, $AX$ bisects $\widehat A$ then $\frac{BX}{CX}=\frac{AB}{AC}$ .
Proof is left as an exercise for the interested reader.

Lemma 2- As in the figure below, extend $IK$ to $J$, such that $\widehat{JEK}=\widehat{IEK}$ . Then $\triangle EJI$ and $\triangle ABC$ are similar and $\frac{JI}{BC}=\frac{EJ}{AB}=\frac{EI}{AC}$.
Proof is somewhat beautiful, therefore it is left as an exercise for the interested reader.

Now for calculations, let us refer to the radius of the in-circle as $r$. Let us also write the lengths of $BC$ , $CA$ and $AB$ as $a$ , $b$ and $c$ , respectively. We have:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{S_{\triangle BIC} + S_{\triangle BIK} + S_{\triangle CIK}}{S_{\triangle ABC}} $$
$$= \frac{a}{a+b+c} + \frac{a.IK}{(a+b+c)r} \qquad (1)$$
To find $IK$ , note that in $\triangle EJI$ , $EK$ bisects $\widehat{JEI}$ . From Lemma 1 we have:
$$\frac{IK}{JK} = \frac{EI}{EJ} $$
$$\therefore \frac{IK}{IJ} = \frac{EI}{EI+EJ} \qquad (2)$$
From Lemma 2 we have:
$$\frac{EI}{EI+EJ} = \frac{b}{b+c} \qquad (3)$$
and
$$\frac{IJ}{EI} = \frac{a}{b} $$
$$\therefore IJ = \frac{a.r}{b} \qquad (4)$$
From (2) , (3) and (4) we can calculate $IK$ :
$$IK = \frac{a.r}{b+c} \qquad (5)$$
Now from (5) and (1) we can calculate the desired ratio of areas:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{a}{a+b+c} (1 + \frac{a}{b+c}) = \frac{a}{b+c}$$
For the particular values of $a=10$ , $b=8$ and $c=7$ we have:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{10}{8+7} = \frac23$$
A: 
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 $\triangle BKC = 5r + \frac{1}{2} (BD+CD) \frac{2r}{3} = \frac{25r}{3}$
$\therefore \frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{2}{3}$.
