Euclidea 4.10

Euclidea 4.10 Square by Adjacent Midpoints
Construct a square, given two midpoints, $${\rm A, B} $$, of adjacent slides.

This is a relatively easy problem. The only challenge is that 7L and 10E are both relatively long solutions. One needs to plan out the whole procedure to stay within the budget. I will not give the proofs as they are both quite straightforward.

Solution 7L



 * 1) Connect $$ \overline{\rm AB} $$;
 * 2) Draw perpendicular from $$ \overline{\rm BC} $$ to $$ \overline{\rm AB} $$;
 * 3) Draw angle bisector of $$ \angle {\rm ABC} $$, and $$ \odot {\rm A} $$ with radius $$ \overline{\rm AB} $$, intersect at $$ {\rm D} $$, let $$ \odot {\rm A} $$ intersect $$ \overline{\rm AB} $$ at $$ {\rm E} $$;
 * 4) Another angle bisector of $$ \angle {\rm DAB} $$, intersect $$ \overline{\rm DB} $$ at $$ {\rm F} $$;
 * 5) From $$ {\rm E} $$, drop a line perpendicular to $$ \overline{\rm AF} $$, intersect $$ \overline{\rm AF} $$ at $$ {\rm G} $$, $$ \odot {\rm A} $$ at $$ {\rm H} $$, and line $$ \overline{\rm CB} $$ at $$ {\rm J} $$;
 * 6) Perpendicular bisector of $$ \overline{\rm HJ} $$ as the last side of the square $$ \overline{\rm KL} $$.

Solution 10E

 * 1) Connect $$ \overline{\rm AB} $$, draw perpendicular bisector of $$ \overline{\rm AB} $$ to find the midpoint $$ {\rm D} $$, and then draw $$ \odot {\rm D} $$, to find points $$ {\rm C} $$ and $$ {\rm E} $$. These are probably the first a few steps that everyone would try if there were no budget limit, suprisingly, they are the right steps, which is a rare case with a bisector used in and E-solutions.
 * 2) Connect $$ \overline{\rm CA} $$, $$ \overline{\rm CB} $$;
 * 3) Draw circle $$ \odot {\rm E} $$, with radius $$ \overline{\rm EC} $$, intersect $$ \overline{\rm AB} $$ and  $$ \overline{\rm AB} $$ at $$ {\rm F, G} $$, and intersect $$ \overline{\rm CD} $$ at $${\rm H} $$;
 * 4) Finish by connecting $$ \overline{\rm FH} $$ and $$ \overline{\rm GH} $$.