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$\angle BCA = 15^{\circ}$

$\angle CBM = \angle ABH$

$\angle BHC = 90^\circ$

Find $|AC|$

enter image description here

The solution states that $\overline{BM}$ is the isogonal conjugate of $\overline{BH}$ but I did not understand this at all. Could you explain this to me or maybe find another way to solve it?

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Isogonal conjugate just means that $\angle CBM = \angle ABH$. There are some properties which may come into play later, but you should be able to read on. – Calvin Lin Feb 4 '14 at 22:14
To check, are you also given that $\angle CBA = 90^\circ$? Otherwise it seems like there could be multiple angles, just by scaling to fit the length of $MH$. – Calvin Lin Feb 4 '14 at 22:17
The solution says that because the median $\overline{BM}$ is the isogonal conjugate of the altitude $\overline{BH}$, $\angle CBA = 90^{\circ}$ but I don't understand that. – Zafer Cesur Feb 4 '14 at 22:26
@CalvinLin It seems he's given $\angle BHC = 90^\circ$, from the drawing. Using that, experimentation in GeoGebra seems to suggest that there is only one case (where $\triangle BMH$ was non-degenerate). – Arthur Feb 4 '14 at 22:26
Oh, yes. $\angle BHC = 90^\circ$ is given. – Zafer Cesur Feb 4 '14 at 22:27
up vote 1 down vote accepted

Call the angle at $A$ $\alpha$ and let $O$ be the circumcenter.

Clearly, $\measuredangle HBA = 90-\alpha$.

Now, by the inscribed angle theorem: $\measuredangle CBO = \frac{180-\measuredangle BOC}2 =90-\measuredangle BAC = 90- \alpha$

Therefore, both $M$ and $O$ lie on the line $BM$ as well as on the perpendicular bisector of $CA$. This means that either the triangle isosceles or $M=O$. In the first case, $M=H$ which is clearly impossible from the given distance.

However, if $M=O$, the triangle has be orthogonal at $B$. The similarity of $ABC$ and $AHB$ now permits to calculate all angles.

You can finally use the sine law and the given distance to calculate $BH$, $AH$ and $CH$ which gives you $AC$.

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$\overline{BM}$ is the Isogonal conjugate of $\overline{BH}$ means BM is pass through the circumcentre as $\overline{BH}$ is pass the orthocentre. so $M$ is circumcentre $\implies \angle CBA= 90^\circ $. now you can find every thing.

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