There must be an identity for this!

Geometry Level 5

$\triangle ABC$ with $AC$ is extended to $D$ . Point $F$ lies on $AB$ , such that $AF : FB = 2 : 3$ . $BC$ and $DF$ intersect at $E$ , such that $CE : BE = 1 : 4$ . If $CD : AC = a: b$ , with $\text{gcd}(a,b) = 1$ and $a$ and $b$ are positive integers, find $a+b$ .

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The answer is 8.

2 solutions

Mark Hennings
Apr 25, 2017

This is an application of Menelaus' Theorem. We have the signed identity $-\frac{a+b}{b} \times \frac{1}{4} \times \frac{3}{2} \; = \; \frac{AD}{DC} \times \frac{CE}{EB} \times \frac{BF}{FA} \; = \; -1$ so that $3(a+b) = 8a$ , and hence $5a = 3b$ , so that $a=3$ , $b=5$ , making the answer $\boxed{8}$ .

Ahmad Saad
Apr 24, 2017

There must be an identity for this!

Try another problem on my set Let's Practice

The answer is 8.

2 solutions

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