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Given is a circle with midpoint $M$ and a chord $AB$ on this circle. $S$ is the intersection of the altitude from $M$ to $AB$. Prove that the locus of all points $S$ is a circle with midpoint $D$ and radius $|AD|$.

Set $\angle A=\alpha$.

$\left.\begin{array}{l}\angle ASM=\angle BSM=90^{\circ} \\ |AM|=|BM| \text{ (circle)} \\ |MS|=|MS| \end{array}\right\}\implies \triangle ASM\cong\triangle BSM$ (RHS) $\implies \angle A=\angle B=\alpha$

$\left.\begin{array}{l} |AB|=2|AS| \\ \angle A=\angle A \\ |AM|=2|AD| \end{array}\right\}\implies \triangle ABM\sim \triangle ASD$ (SAS)

Now I know I have to prove that $DS\parallel MB$ and therefore $\angle B=\alpha=\angle ASD$. However, the proof in my book assumes this without proving. How would I go about completing the proof?

enter image description here

jandewaal
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2 Answers2

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$ASB$ is a right triangle, hence $DA=DS=DM$. This gives that $S$ always belongs to the circle $\Gamma$ having $AB$ as a diamater: now you just have to prove the converse statement. If $S$ belongs to such a circle, $AS=SB$, since an homothethy with center $A$ and factor $2$ sends $\Gamma$ into the original circle.

Jack D'Aurizio
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ASM is a right triangle,
D is the midpoint of AM,

Therefore DA = DS = DM.
And so S ∈ circle with radius |DA|.

Here is an animation to visualize the circle: the locus of midpoint of a chord

But I don't know how to prove that S can be everywhere on the circle.
Maybe you can do that with ∠A ? It seems to make a full turn when A or B moves.

CharlesBoury
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