Euclidea 6.6

Euclidea 6.6 Translate Segment
Construct a segment from the given point $${\rm C} $$ parallel and equal to the given segment $$ \overline{\rm AB} $$

This is a very interesting little problem, using symmetry and a special way to construct parallel lines that is different from those in chapter 5. We only present the 6E solution here.

Solution 6E



 * 1) Draw circle $$ \odot{\rm A} $$ and $$ \odot{\rm B} $$, with radius $$ \overline{\rm AB} $$, just like making a bisector of $$ \overline{\rm AB} $$, get two intersecting points  $$ \odot{\rm D, E} $$;
 * 2) Draw circle $$ \odot{\rm E} $$ with radius $$ \overline{\rm EC} $$;
 * 3) Draw circle $$ \odot{\rm D} $$ with radius $$ \overline{\rm DC} $$, intersect $$ \odot{\rm E} $$ at $$ {\rm F} $$, connect line $$ \overline{\rm CF} $$;

Note at this time, everything is symmetric around the bisector of $$ \overline{\rm AB} $$, which is not drawn. The immediate fact is that $$ \overline{\rm CF} $$ is parallel to $$ \overline{\rm AB} $$.



4. Now draw circle $$ \odot{\rm A} $$ with radius $$ \overline{\rm AF} $$, intersect with line $$ \overline{\rm CF} $$ at point $$ \odot{\rm G} $$, $$ \overline{\rm CG} $$ is the solution.

To see why this is true, again using the symmetry, we know that $$ {\rm ABCF} $$ is an isosceles trapezoid, with $$ \angle{\rm FAB} = \angle{\rm CBA}$$. The circle $$ \odot{\rm A} $$ in the last step, makes another isosceles triangle $$ \triangle{\rm AFG} $$, with $$ \angle{\rm AFG} = \angle{\rm AGF} = \angle{\rm FAB}$$. This makes $$ \overline{\rm AG} \parallel \overline{\rm BC} $$, and $$ {\rm ABCG} $$ a parallelogram, and solves the problem.