4.13. On the sides of triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken, dividing its sides in the ratios $B A_{1}: A_{1} C=p, C B_{1}: B_{1} A=q$ and $A C_{1}: C_{1} B=$ $=r$. The intersection points of segments $A A_{1}, B B_{1}$, and $C C_{1}$ are located as shown in Fig. 4.3. Find the ratio of the areas of triangles $P Q R$ and $A B C$.

## §3. Areas of triangles into which a quadrilateral is divided

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4.13. Using the result of problem 1.3, it is easy to verify that

$$
\frac{B Q}{B B_{1}}=\frac{p+p q}{1+p+p q}, \quad \frac{B_{1} R}{B B_{1}}=\frac{q r}{1+q+q r}, \quad \frac{C R}{C C_{1}}=\frac{q+q r}{1+q+q r}, \quad \frac{C P}{C C_{1}}=\frac{p r}{1+r+p r}
$$

It is also clear that

$$
\frac{S_{P Q R}}{S_{R B_{1} C}}=\frac{Q R}{R B_{1}} \cdot \frac{P R}{R C} \quad \text { and } \quad \frac{S_{R B_{1} C}}{S_{A B C}}=\frac{B_{1} C}{A C} \cdot \frac{B_{1} R}{B B_{1}}
$$

Therefore,

$$
\frac{S_{P Q R}}{S_{A B C}}=\frac{Q R}{B B_{1}} \cdot \frac{P R}{R C} \cdot \frac{B_{1} C}{A C}=\frac{Q R}{B B_{1}} \cdot \frac{P R}{C C_{1}} \cdot \frac{C C_{1}}{C R} \cdot \frac{B_{1} C}{A C}
$$

Considering that

$$
\frac{Q R}{B B_{1}}=1-\frac{p+p q}{1+p+p q}-\frac{q r}{1+q+r q}=\frac{1}{1+p+p q}-\frac{r q}{1+q+r q}=\frac{(1+q)(1-p q r)}{(1+p+p q)(1+q+q r)}
$$

and

$$
\frac{P R}{C C_{1}}=\frac{(1+r)(1-p q r)}{(1+q+q r)(1+r+p r)}
$$

we obtain

$$
\frac{S_{P Q R}}{S_{A B C}}=\frac{(1-p q r)^{2}}{(1+p+p q)(1+q+q r)(1+r+p r)}
$$