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$\triangle ABC$ is isosceles with legs $AC=BC=b$ and angle $\measuredangle ACB=\gamma$. Let $CD$ and $BM$ be altitudes that intersect at $H$. Find $MD$.

enter image description here

I have been struggling with this problem for an hour now. I really don't know how to start. I was thinking about using cosine or sine rule, but it seems useless at the end. For example $$AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\gamma=2b^2-2b^2\cos\gamma$$

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  • $\begingroup$ the angle MBA is known, you can get the point M as the intersection of straight lines whose angular coefficients you know. at that point you just need to make the distance between two points of the plane between D and M $\endgroup$ – Patrick Danzi Jan 21 at 8:46
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Hint: Show that quadrilateral MDBC is inscribed. Then, show that $\angle MAD=\angle AMD$ so $\triangle ADM$ is isosceles with $MD=AD$. Finally, you have to express $AD$ in terms of $b$ and $\gamma$.

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  • $\begingroup$ Can you give me a hint on how to show $\angle MAD=\angle AMD$? $\endgroup$ – Katherine Jan 21 at 8:47
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    $\begingroup$ @nicoledobreva $\angle AMD=180^{\circ}-\angle CMD$. Also, your drawing has $\angle ACD=\gamma$ and not $\angle ACB$. That's a mistake, I guess? $\endgroup$ – bjorn93 Jan 21 at 8:50
  • $\begingroup$ I am trying to find $AB$ using the cosine rule on triangle ABC. I got that $AB=\sqrt{2b^2(1-\cos\gamma)}$. The possible answers are $b\cos\gamma;b\sin\gamma,b\sin\dfrac{\gamma}{2}$ and $b$. Well, using the fact that $MD=AB=\dfrac{1}{2}AB$ we cannot get such answer. $\endgroup$ – Katherine Jan 21 at 8:59
  • $\begingroup$ I meant $MD=AD=..$ $\endgroup$ – Katherine Jan 21 at 9:05
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    $\begingroup$ @nicoledobreva There's no contradiction because you haven't simplified this: $2(1-\cos\gamma)=\sin^2(\gamma/2)$. Then take roots. But you can find $AD$ directly from $\triangle ACD$ using the definition of sine. There's really no need to use law of cosines. $\endgroup$ – bjorn93 Jan 21 at 9:06
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enter image description here

$$\therefore MD = AD = b\sin \gamma$$

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  • $\begingroup$ A really nice diagram. How did you do it? $\endgroup$ – Katherine Jan 21 at 9:43
  • $\begingroup$ @nicoledobreva On Geogebra website and exported image to MS Paint. $\endgroup$ – cosmo5 Jan 21 at 9:45
  • $\begingroup$ I don't see this option for exporting. I have GeoGebra file, PNG image, SVG image, PDF document, 3D Print. Which one do you choose? $\endgroup$ – Katherine Jan 21 at 9:46
  • $\begingroup$ Oh, I see. And then you exported in MS Paint? Can I ask why? $\endgroup$ – Katherine Jan 21 at 9:48
  • $\begingroup$ @nicoledobreva To crop the image so it appears big here. You can upvote if you found this answer useful. $\endgroup$ – cosmo5 Jan 21 at 9:49

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