Finding Locus of a Midpoint of a Chord

Question

Let $A$ and $B$ be points, and let $C$ be a point that varies on the open semicircle with diameter $AB$. Construct squares externally on sides $AC$ and $BC$, and let $D$ and $E$ be the centers of these squares, respectively. Find the locus of the midpoint $M$ of $DE$.

I found that the locus of $M$ is the upper half of the inscribed circle in the semicircle with $AB$ as the diameter, however, I am having trouble proving it, and proving that there are no other places where $M$ can exist. Can I have a hint as to how to prove it geometrically? Thanks!

Answer 1

Hints:

Let $O$ the midpoint of $AB$ and $Y$ the point where line $DC$ intercepts the circle$(O, OA)$.

1. Show that $C$ is a point of $DE$.

2. Show that $m(\angle AOY)=90 ^{\circ}$. Use the inscribed angle theorem.

3. Show that $m(\angle DMO)=90 ^{\circ}$.

So, if $Y$ and $O$ are fixed points and $OMY$ is a right-angled triangle, what comes after?

Answer 2

Hint:-

If we use $A$, $B$ as $(x_i,y_i)$ solution will become a bit complicated so you can try solving it by taking $A$ $(-1,0)$ and $B$ $(1,0)$.

Let the point $C$ be $(\cos x,\sin x)$.

For finding center of the square formed by joining $AC$, we will first have to find their mid point and shift it up by $\dfrac{AC}{2}$ on their $\perp$ bisector. Midpoint of $AC$ is $\left(\dfrac{\cos x-1}{2},\dfrac{\sin x}{2}\right)$, and by using distance form you can easily find the midpoint of the square and then you just have to eliminate $\sin, \cos$ by using the formula $\sin^2x+\cos^2x=1$.