Problem: A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_{1}$, $t_{2}$, and $t_{3}$ in the [figure](A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_{1}$, $t_{2}$, and $t_{3}$ in the figure, have areas $4$, $9$, and $49$, respectively. Find the area of $\triangle ABC$. ), have areas $4$, $9$, and $49$, respectively. Find the area of $\triangle ABC$. Solution: It is not bad to see $t_1, t_2, t_3$ are similar. Further the reatio of their areas is equal to the ratio of their side lengths. Let $a,b,c$ be the side lengths of triangle $t_1$ (following assignment clockwise). We see the side lengths of $t_2$ and $t_3$ are hence $\frac{3}{2}a, \frac{3}{2}b, \frac{3}{2}c,$ and $\frac{7}{2}a,\frac{7}{2}b,\frac{7}{2}c$ respectively. Finally, note the side length of the grade triangle is equal to the sum of the side lengths of the smaller triangles, so we find $A = \frac{(a+\frac{3}{2}a+\frac{7}{2}a)(b+\frac{3}{2}b+\frac{7}{2}b)\sin \theta}{2} = 4\cdot 36 = 144.$