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if $M$ is the midpoint of $BC$ then find the measure of $AMC$.

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I tried to use the angles to find $AMC$ but I don't know how to use that $M$ is the midpoint of $BC$.

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I asked this from our teacher and he give this answer may helps also this answer don't uses trigonometry:

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$BNC=90$degrees and $BAN=45$ then we have $AN=BN$ also because $BNC$ is right angle $MN=BM$ also because $NBM=60$,the triangle $BNM$ is equilateral.Also $MNA=30$ then because $MN=AN$ the angle $NMA=75$ then we have $AMC=45$.

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  • $\begingroup$ What is your question? You want to know why the angle is 45°? (do you want a proof... do you want more explanation on the proof provided by your teacher?) Or...? $\endgroup$ Jun 10, 2016 at 17:58
  • $\begingroup$ No I just added an answer. $\endgroup$ Jun 11, 2016 at 5:30
  • $\begingroup$ add answer to which quesstion!? the asker's name and the answerer's name are same... ☺ $\endgroup$ Jun 11, 2016 at 7:22
  • $\begingroup$ Ok I first asked the question before then tommorow our teacher give this answer I put it because I think it is usuful $\endgroup$ Jun 11, 2016 at 7:36
  • $\begingroup$ OK!... then... I wrote a fully explained... but short... without talking about unnecessary facts... $\endgroup$ Jun 11, 2016 at 12:07
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The answer is $\color{red}{45^\circ}$. You may embed the construction in the complex plane by assuming $C=1, B=-1$. Then $A$ has to fulfill: $$A-C = \lambda e^{5\pi i/6},\qquad A-B = \mu e^{\pi i/12},\qquad (\lambda,\mu\in\mathbb{R}) $$ but that leads to $A=\eta(1+i)$, and the claim follows.


An alternative, elementary way. By assuming $BM=MC=1$ we have: $$ AC = 2\frac{\sin 15^\circ}{\sin 135^\circ}=-1+\sqrt{3},\qquad AB = 2\frac{\sin 30^\circ}{\sin 135^\circ}=\sqrt{2}$$ by the sine theorem. By Stewart's theorem we have: $$ AM^2 = \frac{2AB^2+2AC^2-BC^2}{4} = \left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)^2$$ and by the sine theorem again: $$ \sin\widehat{CMA} = AC\cdot\frac{\sin 30^\circ}{AM} = \frac{1}{\sqrt{2}},$$ from which $\widehat{CMA}=\color{red}{45^\circ}$.

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  • $\begingroup$ How can you directly conclude that $A=\eta(1+i)$ $\endgroup$
    – Hamza
    Jun 9, 2016 at 17:28
  • $\begingroup$ You have to make some computation, but that is an easy task. $\endgroup$ Jun 9, 2016 at 17:31
  • $\begingroup$ But isn't it better to use geometry to solve a geometry question? Analytical methods seem non intuitive $\endgroup$
    – N.S.JOHN
    Aug 14, 2016 at 12:40
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Let $X$ be the point on $BC$ exactly below $A$ and $MX=z$ for comfort.

$BX=\frac{BC}{2}+z$ and $XC=\frac{BC}{2}-z$

Now:

$$\tan(\alpha )= \frac{AX}{BX}\implies \tan(\alpha )= \frac{AX}{\frac{BC}{2}+z}$$

$$\tan(AMC )= \frac{AX}{Z}$$

$$\tan(\beta )= \frac{AX}{XC} \implies \tan(\beta )= \frac{AX}{\frac{BC}{2}-z}$$

Now you have 3 questions with 3 variables $z,AX,\tan(AMC)$ and the rest is given. Let $AMC= \gamma$

$$AX=z \tan(\gamma)$$

$$(1)\tan(\alpha )= \frac{z \tan(\gamma)}{\frac{BC}{2}+z}\implies \frac{BC\tan(\alpha )}{2}=z(\tan(\gamma)-\tan(\alpha ))$$

$$(2)\tan(\beta )= \frac{z \tan(\gamma)}{\frac{BC}{2}+z}\implies \frac{BC\tan(\beta )}{2}=z(\tan(\gamma)+\tan(\beta ))$$

And by dividing (1)/(2) we get:

$$\frac{\tan(\alpha )}{\tan(\beta )}=\frac{\tan(\gamma)-\tan(\alpha )}{\tan(\gamma)+\tan(\beta )}$$

Hence: $$\tan(\gamma)=\frac{2\tan(\alpha)\tan(\beta)}{\tan(\beta)-\tan(\alpha)}$$

$$\gamma=\arctan(\frac{2\tan(30^\circ)\tan(15^\circ)}{\tan(30^\circ)-\tan(15^\circ)})$$

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  • $\begingroup$ now whats the answer? $\endgroup$ Jun 9, 2016 at 14:37
  • $\begingroup$ I think I can safely let you proceed.. You have $\alpha$ and $\beta$. All you need is a calculator. $\endgroup$
    – havakok
    Jun 9, 2016 at 14:57
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enter image description here Continue $CA$, draw the ray $\vec{CA}$. Then draw $BN$, in which it is perpendicular to $\vec{CA}$.

$\triangle BCN$ is a 30-60-90 triangle. as we know about this triangle. $BN=\frac12 BC$. So $BN=BM$. This tells us $\triangle ABN$, is isosceles, so $\angle BMN=\angle BNM=\frac{180^\circ-\angle MBN}2=\frac{120^\circ}{2}=60^\circ$.

  • Now we know $\triangle BMN$ is equilateral.

Therefor $$MN=BN=BM\qquad(1)\\ \text{and }\angle MBN=60^\circ\qquad{ }$$

By this, we get $\angle CMN=180^\circ-\angle BMN=180^\circ-60^\circ$, $$\text{therefor }\angle CMN=120^\circ\ (2)$$

$\angle ABN=\angle CBN-\angle CBA=60^\circ-15^\circ=45^\circ$. So, the other angle of the right triangle $\triangle ABN$, i.e. $\angle BAN$, should be $180^\circ-90^\circ-45^\circ=45^\circ{}^\dagger$, therefor $\triangle ABN$ is isosceles.

So we would have $AN=BN$. By (1) we get $MN=AN$, so $\triangle AMN$ is isosceles. Therefor $$\angle AMN=\angle MAN=\frac{180^\circ-\angle ANM}2=\frac{180^\circ-30^\circ}2=75^\circ$$

Now, by using (2) we can find the size of $\angle AMC$, which is $$\angle AMC=\angle CMN-\angle AMN=120^\circ-75^\circ=45^\circ$$

${}^\dagger$let's note that one may also know that $\angle BAN$ is equal to $45^\circ$, since it is an external angle of the triangle $\triangle ABC$

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Another geometric solution: enter image description here Let $E$ be the foot of the altitude through $A$. Let $D$ be the midpoint of $\overline{AC}$, so $\triangle EDC$ is isosceles with $\angle DEC=\angle ECD=30$ and $ED = DC$.

Both $M$ and $D$ are midpoints, so $\triangle ABC\sim\triangle DMC$. Conclude $\angle DMC=\angle ABC=15$. Knowing $\angle DEC=30$, compute $\angle EDM=15$ so $\triangle MED$ is isosceles. Hence $ME=ED$.

Since $D$ is the midpoint of $AC$, we know $DC=\frac12 AC$. But $\triangle AEC$ is a 30-60-90 triangle so also $AE=\frac12 AC$.

Putting it all together, we've shown $ME=ED=DC=\frac12 AC=AE$. So $\triangle MEA$ is isosceles. Since $\angle MEA$ is right, it follows that $\angle AME=45$.

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