In triangle $ABC$, we are given $AB=18$, $AC=12$, and $BC=15$. Point $D$ lies on $AB$ such that $BD=6$. A line parallel to $BC$ through $D$ intersects $AC$ at $E$. We want to find the area of quadrilateral $BDEC$ (denoted as $SADE$ in the text, likely a typo). We are also told that $D$ lies on $AF$, where $AF$ is the altitude from $A$ to $BC$.

1. Problem Description

In triangle $ABC$, we are given $AB=18$, $AC=12$, and $BC=15$. Point $D$ lies on $AB$ such that $BD=6$. A line parallel to $BC$ through $D$ intersects $AC$ at $E$. We want to find the area of quadrilateral $BDEC$ (denoted as $SADE$ in the text, likely a typo). We are also told that $D$ lies on $AF$, where $AF$ is the altitude from $A$ to $BC$.

2. Solution Steps

Since $DE$ is parallel to $BC$, triangle $ADE$ is similar to triangle $ABC$. We have $AD = AB - BD = 18 - 6 = 12$.

The ratio of similarity between triangles $ADE$ and $ABC$ is $\frac{AD}{AB} = \frac{12}{18} = \frac{2}{3}$.

Therefore, $\frac{AE}{AC} = \frac{DE}{BC} = \frac{2}{3}$.

$AE = \frac{2}{3}AC = \frac{2}{3}(12) = 8$.

$DE = \frac{2}{3}BC = \frac{2}{3}(15) = 10$.

Let $h$ be the height of triangle $ABC$ with base $BC$, and let $h'$ be the height of triangle $ADE$ with base $DE$. Since triangles $ADE$ and $ABC$ are similar, we have $\frac{h'}{h} = \frac{2}{3}$, so $h' = \frac{2}{3}h$.

We know that the area of triangle $ABC$ can be written as $\frac{1}{2}BC \cdot h = \frac{1}{2}(15)h = \frac{15}{2}h$.

The area of triangle $ADE$ is $\frac{1}{2}DE \cdot h' = \frac{1}{2}(10)(\frac{2}{3}h) = \frac{10}{3}h$.

The area of quadrilateral $BDEC$ is the area of triangle $ABC$ minus the area of triangle $ADE$, which is

$\frac{15}{2}h - \frac{10}{3}h = \frac{45 - 20}{6}h = \frac{25}{6}h$.

We need to find $h$. Let $a = 15$, $b = 12$, $c = 18$.

Let $s = \frac{a+b+c}{2} = \frac{15+12+18}{2} = \frac{45}{2}$.

By Heron's formula, the area of triangle $ABC$ is $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{45}{2}(\frac{45}{2}-15)(\frac{45}{2}-12)(\frac{45}{2}-18)} = \sqrt{\frac{45}{2}(\frac{15}{2})(\frac{21}{2})(\frac{9}{2})} = \sqrt{\frac{45 \cdot 15 \cdot 21 \cdot 9}{16}} = \frac{1}{4}\sqrt{3^2 \cdot 5 \cdot 3 \cdot 5 \cdot 3 \cdot 7 \cdot 3^2} = \frac{1}{4}\sqrt{3^6 \cdot 5^2 \cdot 7} = \frac{3^3 \cdot 5}{4}\sqrt{7} = \frac{27 \cdot 5}{4}\sqrt{7} = \frac{135}{4}\sqrt{7}$.

The area of triangle $ABC$ is also equal to $\frac{1}{2} \cdot BC \cdot h = \frac{15}{2} h$.

Therefore, $\frac{15}{2}h = \frac{135}{4}\sqrt{7}$, which means $h = \frac{135}{4}\sqrt{7} \cdot \frac{2}{15} = \frac{270}{60}\sqrt{7} = \frac{9}{2}\sqrt{7}$.

The area of quadrilateral $BDEC$ is $\frac{25}{6}h = \frac{25}{6} \cdot \frac{9}{2}\sqrt{7} = \frac{225}{12}\sqrt{7} = \frac{75}{4}\sqrt{7}$.

3. Final Answer

The area of quadrilateral $BDEC$ is $\frac{75\sqrt{7}}{4}$.