Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.7 National olympiad, round 2 Prove it

Example 6 Given a line segment $A B$ and a line $C D$ parallel to $A B$ , let $n$ be a positive integer, $n \geqslant 2$ . Using only a straightedge, construct a point that divides $A B$ into $n$ equal parts.

Construct a point that divides $A B$ into $n$ equal parts using only a straightedge, given a line segment $A B$ and a line $C D$ parallel to $A B$ , where $n$ is a positive integer, $n \geqslant 2$ .

Solution

Proof: By mathematical induction. When $n=2$ , the proposition holds. Suppose point $I$ is the $(n-1)$ -th equal division point of $AB$ (closest to point $B$ ) for $n \geqslant 3$ . Connect $IG$ to intersect $BF$ at $J$ , and extend $EJ$ to intersect $AB$ at $K$ . Then,
$\frac{AK}{BK} = \frac{\triangle AEJ}{\triangle BEJ} = \frac{\triangle AEJ}{\triangle AFJ} \cdot \frac{\triangle AFJ}{\triangle ABJ} \cdot \frac{\triangle AJB}{\triangle EBJ} = \frac{AE}{AF} \cdot \frac{FJ}{BJ} \cdot \frac{AF}{EF} = \frac{\triangle AGE}{\triangle AGF}.$
$\frac{\triangle FIG}{\triangle BIG} \cdot \frac{AF}{EF} = \frac{\triangle AGE}{\triangle AGF} \cdot (n-1) \frac{\triangle FAG}{\triangle BAG} \cdot \frac{AF}{EF} = (n-1) \frac{EG}{BG} \cdot \frac{AF}{EF} = (n-1) \frac{\triangle EFG}{\triangle BFG} \cdot \frac{\triangle AFG}{\triangle EFG} = n-1,$
so point $K$ is the $n$ -th equal division point of $AB$ (closest to point $B$ ).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.