$\left[\begin{array}{c}\text { [Ratio of areas of triangles with a common angle] } \\ {[\underline{\text { Heron's Formula }}]}\end{array}\right.$

Given a triangle $ABC$ with sides $AB=6$, $AC=4$, and $BC=8$. Point $D$ lies on side $AB$, and point $E$ lies on side $AC$, such that $AD=2$ and $AE=3$. Find the area of triangle $ADE$.

#
$$
S_{\triangle \mathrm{ADE}}=\frac{A D}{A B} \cdot \frac{A E}{A C} \cdot S_{\triangle \mathrm{ABC}}
$$

## Solution

Let $p$ be the semi-perimeter of triangle $ABC$. From the problem, we have $p=\frac{1}{2}(AB+AC+BC)=\frac{1}{2}(6+4+8)=9$. By Heron's formula,

$$
s_{\triangle \mathrm{ABC}}=\sqrt{p(p-AB)(p-AC)(p-BC)}=\sqrt{9 \cdot 3 \cdot 5 \cdot 1}=3 \sqrt{15}
$$

$$
S_{\triangle \mathrm{ADE}}=\frac{A D}{A B} \cdot \frac{A E}{A C} \cdot S_{\triangle \mathrm{ABC}}=\frac{2}{6} \cdot \frac{3}{4} \cdot 3 \sqrt{15}=\frac{3 \sqrt{15}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-28.jpg?height=292&width=688&top_left_y=234&top_left_x=685)

Answer

$\frac{3 \sqrt{15}}{4}$.