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In the Locus 10, I get it till the point when it says that given triangles AOM and BOM, we have $$AM^2\ =\ OM^2+OA^2-2OA*OD$$And the same for $BM^2$, where does the author come to this conclusions??, I've been trying to understand this for some couple of hours. Thanks.

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The quick way to do this is to use the cos rule, as Paul Aljabar mentions. But if you don't know the cos rule you can do it using Pythagoras' theorem, essentially proving the cos rule.

We want to show $$AM^2 = OM^2 + OA^2 - 2OA \cdot OD$$ We have two right triangles, $MDA$ and $MDO$. Thus $$MD^2 = AM^2 - AD^2 = OM^2 - OD^2$$ So $$\begin{align} AM^2 & = OM^2 + AD^2- OD^2\\ & = OM^2 + AD^2 + OD^2 + 2AD \cdot OD - 2OD^2 - 2AD \cdot OD\\ & = OM^2 + (AD + OD)^2 - 2(OD +AD) \cdot OD\\ & = OM^2 + OA^2 - 2OA \cdot OD\\ \end{align}$$

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we have by the Theorem of cosines $$AM^2=OM^2+OA^2-2OM\cdot OA\cos(\angle AOM)$$ further we have $$\cos(\angle AOM)=\frac{OD}{OM}$$

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For fun:

Polar coordinates and trigonometry.

Consider a circle with radius $1$ centred at $(0,0).$

Choose $2$ points on the circle:

$A (\cos\alpha,\sin\alpha)$, and $B(\cos\beta,\sin\beta)$.

Let $\alpha > \beta$. The points $A$ and $B$ on the circle specify a chord of a certain length determined by $\alpha$ and $\beta$.

Midpoint $M$ of $AB$:

$1/2(\cos\alpha +\cos\beta, \sin\alpha +\sin\beta).$

Squared distance of $M$ to origin:

$r^2:= (1/4)×$

$[(\cos\alpha +\cos\beta)^2+(\sin\alpha +\sin\beta)^2]$;

$r^2= (1/4)×$

$(2+ 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) $;

$r^2= (1/4)(2+ 2\cos(\alpha-\beta)).$

Locus of $M:$

Keeping $\alpha-\beta$ constant, I.e. considering chords of a specific length, we move around the circle by varying $\alpha$.

Since $r$, the.distance $OM$, depends only on the difference $\alpha -\beta$, we find that the locus of $M$ lies on a concentric circle with radius r.

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