Problem:
A point P is chosen in the interior of △ABC so that when lines are drawn through P parallel to the sides of △ABC, the resulting smaller triangles, t1​,t2​ and t3​ in the figure, have areas 4,9 and 49, respectively. Find the area of △ABC.
Solution:
Let T denote the area of △ABC, and denote by T1​,T2​ and T3​ the areas of t1​,t2​ and t3​, respectively. Moreover, let c be the length of AB, and let c1​,c2​ and c3​ be the lengths of the bases parallel to AB of t1​,t2​ and t3​, respectively. Then, in view of the similarity of the four triangles, one has
T​T1​​​=cc1​​,T​T2​​​=cc2​​ and T​T3​​​=cc3​​
Moreover, since c1​+c2​+c3​=c, it follows that
