Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.6 AMC 10/12 Find the answer South Africa

$D$ is a point on side $AB$ of $\triangle ABC$ , $E$ is a point on $CD$ and $F$ is a point on $CE$ . The areas of triangles $AED$ , $AEC$ , $BFD$ and $BFC$ are $6$ , $10$ , $17$ and $7$ , respectively. What is the area of $\triangle BEF$ ?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

If two triangles have the same height, then the ratio of their areas is equal to the ratio of their bases. It follows that $\frac{DE}{EC} = \frac{6}{10} = \frac{3}{5}$ , so $\frac{DE}{DC} = \frac{3}{3+5} = \frac{3}{8}$ . Similarly, $\frac{DF}{FC} = \frac{17}{7}$ , so $\frac{DF}{DC} = \frac{17}{17+7} = \frac{17}{24}$ . Therefore $\frac{EF}{DC} = \frac{DF - DE}{DC} = \frac{17}{24} - \frac{3}{8} = \frac{17-9}{24} = \frac{8}{24} = \frac{1}{3}$ . It follows that $\triangle BEF = \frac{1}{3}\triangle BDC = \frac{1}{3}(17+7) = 8$ .

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