Curious Chord in a Circle

Let $N$ be the feet of the perpendicular line to $AB$ passing through $M$, and let $L$ be the feet of the perpendicular line to $BC$ passing through $M$. First we will make some observations.

For instance, $N, D, L$ are collinear because of the Simson line theorem. Since $M$ is the midpoint of the minor arc $AB$, and $C$ lies in te minor arc $BM$, we have $CM$ is the external angle bisector o $\angle BCA$. Then, we may notice the cuadrilateral $MLCD$ is cyclic, because $MD$ and $ML$ are perpendicular to $AC$ and $BC$ respectively, but also, $MC$ is the internal angle bisector of $\angle LCD$, so we have triangles $MDC$ and $MLC$ are congruent, and thus, $CD=CL$.

Now, it follows by Menelao's theorem that (in absolute value) we have

$AN\cdot BL\cdot CD=BN\cdot CL\cdot AD$

but, since $M$ is midpoint of arc $AB$, we have $AN=BN$, and we have $CD=CL$, so it follows that $AD=BL$ so we have

$45=15+CL=15+CD$

so we conclude that $CD=30$.

AM=MB because they are on the same arc. \angle MBC = \angle MAC because they are on the same arc, so we can call them \alpha. For Carnot theorem we have MB^2+BC^2-2 BC MB \cos \alpha=AM^2+AC^2-2 AC AM \cos \alpha. We can see that \cos \alpha=\frac {AD}{AM} from the triangle ADM, so we have AM^2+BC^2-2 BC AM \frac {AD}{AM}=AM^2+AC^2-2 AC AM \frac {AD}{AM}, and then BC^2-2 BC AD=AC^2-2 AC AD. Doing calculations and substituting values of AD and BC, we obtain 225-1350=(DC+45)^2-90(DC+45). Finally we have DC^2=900, so DC=30.

Archimedes Borden Chord Theorem states that for a broken chord, AC, CB, the feet of perpendicular from the center of the minor arc AB onto the broken chord bisects the entire broken chord.

Hence AD is half the length of the entire broken chord (AC+BC).

Hence CD = AD-BC = 30

Let $P$ be the point on $BC$ such that $MP \perp BC$ . Connect lines $MP, MC, AM, BM$ . Now since A, M, C, B are concyclic, we have $\angle MAD=\angle MAC=\angle MBC=\angle MBP$ . And $M$ is the midpoint of arc $AB$ implies $AM=BM$ , we have that $\Delta MAD \cong \Delta MBP$ . So $MD=MP$ . Thus $\Delta MDC \cong \Delta MPC$ by HL congruence. As a result, $PC=DC$ , so $DC=PC=PB-CB=AD-CB=30$ .

We will show that $AD = BC + CD$ .

Place a point $E$ on the segment $AD$ such that $AE = BC$ . Since $M$ is the midpoint of arc $\stackrel{\frown}{AB}$ , thus $AM = MB$ . We also have that $\angle MAE = \angle MBC$ as they subtend the same arc. Thus, triangles $MAE$ and $MBC$ are congruent by side-angle-side. Therefore $ME = MC$ and so $MEC$ is an isosceles triangle. Since $MD$ is perpendicular to $EC$ and $MEC$ is isosceles, thus $ED = CD$ . Hence $AD = AE + ED = BC + CD$ .

So, we have $CD = AD - BC = 45 - 15 = 30$ .

Let $x= AD$ , $y=BC$ , $z=AM=MB$ , $\theta=\angle MAD=\angle CBM$ . From triangle $ADM$ , $\cos \theta = \frac {x}{z}$ . From triangle $MBC$ , $\cos \theta = \frac {z}{2y}$ . Therefore, $z^2=2xy$ (1) and using pythagorean theorem, $MD^2=z^2-x^2=y^2-DC^2$ (2) Substituting (1) to (2), $2xy-x^2=y^2-DC^2, DC^2=(x-y)^2$ (3) Substituting $x=45$ and $y=15$ to (3) $DC^2=(45-15)^2, DC=30$ .

[Latex edits, Punctuation edits - Calvin]