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


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:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{10}{8+7} = \frac23$$
