In the triangle $\triangle A B C, M$ is the midpoint of the side $A C, D$ is a point on the side $B C$, such that $A D$ is the bisector of the angle $B \widehat{A} C$, and $P$ is the intersection point of $A D$ and $B M$. Knowing that $A B=10 \mathrm{~cm}, A C=30 \mathrm{~cm}$ and the area of the triangle $\triangle A B C$ measures $100 \mathrm{~cm}^{2}$, calculate the area of the triangle $\triangle A B P$.
The triangles $\triangle A B M$ and $\triangle A B C$ have the same height $d$ relative to their respective bases $A M$ and $A C$. Since $M$ is the midpoint of $A C$, we obtain

$$
\frac{\text { area }(\triangle A B M)}{\text { area }(\triangle A B C)}=\frac{\frac{1}{2} A M \cdot d}{\frac{1}{2} A C \cdot d}=\frac{A M}{A C}=\frac{1}{2}
$$

so that

$$
\text { area }(\triangle A B M)=\frac{1}{2} \text { area }(\triangle A B C)=\frac{1}{2} \quad 100=50 \mathrm{~cm}^{2} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-096.jpg?height=462&width=805&top_left_y=2236&top_left_x=660)

Analogously,

$$
\frac{\text { area }(\triangle A B P)}{\text { area }(\triangle A B M)}=\frac{B P}{B M} \text {. }
$$

By the Angle Bisector Theorem,

$$
\frac{B P}{P M}=\frac{A B}{A M}=\frac{10}{15}=\frac{2}{3}
$$

thus, $P M=\frac{3}{2} B P$. Then we obtain

$$
\frac{\text { area }(\triangle A B P)}{\text { area }(\triangle A B M)}=\frac{B P}{B M}=\frac{B P}{B P+P M}=\frac{B P}{B P+\frac{3}{2} B P}=\frac{B P}{\frac{5}{2} B P}=\frac{2}{5}
$$

so that area $(\triangle A B P)=\frac{2}{5}$ area $(\triangle A B M)=\frac{2}{5} \cdot 50=20 \mathrm{~cm}^{2}$.