The Persistent Angle

This solution may not be the best approach but is the only one I can think of right now.

A little bit background info first. An isosceles triangle with $108°$ or $36°$ vertex angle is rather special. The one with $108°$ has sides of ratio $1:1:\frac{1+\sqrt{5}}{2}$ , while the one with $36°$ as the vertex angle has sides of ratio $\frac{1+\sqrt{5}}{2}:\frac{1+\sqrt{5}}{2}:1$ .

So here's my solution:

First, extend $CM$ to $P$ , making $CM=MP$ . $AP$ and $BC$ intersects at $Q$ .

$BD$ and $CP$ are bisecting each other, thus $BCDP$ is a parallelogram.

Therefore, $\overline{\rm AD}\|\overline{\rm BP}$ and $CD=BP$ .

Then due to the parallel lines, $\frac{BP}{BQ}=\frac{AC}{QC}$ .

As given, $BC=AD$ , and $CD=BP$ , we can conclude that $BP+AC=BQ+QC$ .

Combining the two equations we get $BP=BQ$ and $AC=QC$ .

$\triangle ABC$ is isosceles and $\angle A$ is measured as $108°$ , which indicates that $\angle ACB$ and $\angle ABC$ are both $36°$ and due to parallelism, $\angle PBQ$ is also $36°$ . Therefore, $\triangle ACQ$ and $\triangle PBQ$ are two isosceles triangles with vertex angles measured as $36°$ , and $\triangle ABC$ is an isosceles triangle with vertex angle measured as $108°$ .

From $\triangle ACQ$ and $\triangle PBQ$ we can get $\frac{BQ}{PQ}=\frac{CQ}{AQ}=\frac{1+\sqrt{5}}{2}$ . Then $\frac{BC}{AP}=\frac{BQ+CQ}{PQ+AQ}=\frac{1+\sqrt{5}}{2}$ .

From $\triangle ABC$ we can get $\frac{BC}{AB}=\frac{BC}{AC}=\frac{1+\sqrt{5}}{2}$ .

Therefore, $AB=AP=AC$ .

Recall that $m\angle ABC=m\angle PBQ=36°$ , we can get $m\angle ABP=72°$ .

As $AB=AP$ , $\triangle ABP$ is an isosceles triangle with $\angle ABP$ measured as $72°$ . Therefore, $m\angle BAP=36°$ , so $m\angle CAP=72°$ . As $AC=AP$ , $\triangle CAP$ is an isosceles triangle with a $72°$ vertex angle, leading to the conclusion that $m\angle ACM=54°$ .

And of course, an easier way is to draw the whole picture precisely and measure with a protractor.

$construct\quad a\quad //gram\quad with\quad AB//ED,AD//BE\\ Let\quad \angle CBM\quad be\quad \theta \\ \angle ADB=\angle DBE=36-\theta \\ BC=AD=BE.\angle BCE=\angle BEC=72\\ \angle ECD=180-36-72=72,\\ \angle CDE=\angle ADB+BDE=36+\theta +36-\theta =72\\ CE=DE=AB=AC,and\quad \angle DEC\quad =36\\ \triangle ACE\quad is\quad an\quad isos.triangle\\ \angle CAE=\angle CEA=36\\ M\quad is\quad also\quad the\quad mid-pt.\quad of\quad AE\\ \angle CME=90\quad and\quad \angle ECM\quad is\quad 54\\ \angle BCM=72-54=18\\ \angle ACM=36+18=54$

Here's my solution. It does not require extending the diagram, but you'll probably need a calculator.

Since $\triangle ABC$ is isosceles, we know that $\angle ABC = \angle ACB = \dfrac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}$

Now, I let the value of $BC$ be $10$ cm (you can use any other value, the answer will be the same)

Next, find $AC$ and $AB$ with sine rule:

$\dfrac{AC}{\sin 36^{\circ}} = \dfrac{10}{\sin 108^{\circ}} \implies AC = AB = 6.18$ cm

Since $AD = BC = 10$ cm, we know that $CD = 10 - 6.18 = 3.82$ cm.

$\angle BCD = 180^{\circ} - 36^{\circ} = 144^{\circ}$

With these values, we can calculate $BD$ with the cosine rule:

$BD^2 = BC^2 + CD^2 - 2(BC)(CD) \cos \angle BCD$

$BD = \sqrt{10^2 + 3.82^2 - 2(10)(3.82)\cos 144^{\circ}} = 13.28$ cm

Use this value obtained to find $\angle ADB$ :

$\dfrac{\sin \angle ADB}{6.18} = \dfrac{\sin 108^{\circ}}{13.28} \implies \angle ADB =26.27^{\circ}$

We know that $BM = MD = \dfrac{13.28}{2} = 6.64$ cm, and $\angle CBM = 180^{\circ} - 108^{\circ} - 36^{\circ} - 26.27^{\circ} = 9.73^{\circ}$

We need to find one more thing: the length of $CM$ :

$CM^2 = BC^2 + BM^2 - 2(BC)(BM) \cos \angle CBM$

$CM = \sqrt{10^2 + 6.64^2 - 2(10)(6.64)\cos 9.73^{\circ}} = 3.63$ cm

With this, we can finally determine $\angle BCM$ :

$\dfrac{\sin \angle BCM}{6.64} = \dfrac{\sin 9.73^{\circ}}{3.63} \implies \angle BCM =18^{\circ}$

Therefore, $\angle ACM = \angle ACB + \angle BCM = 36^{\circ} + 18^{\circ} = \boxed{54^{\circ}}$

CONSTRUCTIOn:-

• Draw a line from B || to CM and intersecting AC at E .

• Now from E draw a line || to BC intersecting AC at F .

Let AB=AC=a and $\angle MCB= \phi$

By applying Sne Rule in $\triangle ABC \Rightarrow \dfrac{BC}{\sin(A)}=\dfrac{AC}{\sin(B)} \Rightarrow BC=2.a . \cos\dfrac{\pi}{5}$

BC =AD = $2a \cos\dfrac{\pi}{5}$

In $\triangle BDE$ and $\triangle MDC$ :-

CM || BE By BPT EC=CD = $2.a .\cos\dfrac{\pi}{5}-a$ $\Rightarrow AE=2a(1-\cos\dfrac{\pi}{5})$ the

In $\triangle ABC$ $\triangle AFE$ FE || BC By BPT $\dfrac{FE}{BC}= \dfrac{AF}{AB} = \dfrac{AE}{AC} \Rightarrow FE = 4.a.\cos\dfrac{\pi}{5}(1-\cos\dfrac{\pi}{5})$ ) and BF=EC

Now Since FE||BC and MC||BE $\angle FEB=\angle MCB=\phi$

By applying Sine Rule in $\triangle BFE \Rightarrow \dfrac{\sin\angle FEB}{\sin\angle FBE}=\dfrac{FB}{FE}$

This gives $\dfrac{\sin\phi}{\sin(\dfrac{\pi}{5}-\phi)}=\dfrac{2a \cos \dfrac{\pi}{5}-a}{4.a.\cos\dfrac{\pi}{5}(1-\cos\dfrac{\pi}{5})}=1$ $\Rightarrow \sin\phi=\sin(\dfrac{\pi}{5}-\phi)$

$\Rightarrow \phi= \dfrac{\pi}{10}$

Note:- BPT is Basic Proportionality Theorem (Thales Theorem).

So the required angle is $18^\circ+36^\circ=54^\circ$

Draw BE // MC, where E belongs to AD. We will prove that BE is bisector of the angle ABC. Let us note m(ACB)=36= $\alpha$ = $\frac {\pi}{5}$ . Thus 0 = sin $\pi$ = sin 5 $\alpha$ . Doing the calculus (expanding the sin 5 $\alpha$ ), we will get the folowing relation: 16 $\sin^4$ $\alpha$ - 20 $\sin^{2}$ $\alpha$ + 5 = 0. (1) In order to prove the BE is the bisector, we will use the theorem of bisector: "If ABC is a triangle then AM is bisector if and only if $\frac{AB}{AC}$ = $\frac{MB}{MC}$ , where M belongs to BC." Then, using the hypothesis and the auxiliary construction, we have: $\frac{BC}{AB}$ = $\frac{AD}{AC}$ = $\frac{AC + CD}{AC}$ = 1 + $\frac{CE}{AC}$ = 1 + $\frac{CE}{AE + CE}$ = 1 + $\frac{1}{1 + x}$ = $\frac{x + 2}{x + 1}$ , where x = $\frac{AE}{CE}$ . Using the sine theorem in ABC, we will have: $\frac{BC}{AB}$ = $\frac{sin3{\alpha}}{sin{\alpha}}$ = 3 - 4 $\sin^2{\alpha}$ = $\frac{x+2}{x+1}$ . Solving the equation in x, we'll get $\frac{1}{x}$ = $\frac{2 - 4 sin^2{\alpha}}{4 sin^2{\alpha} - 1}$ . But, $\frac{2 - 4 sin^2{\alpha}}{4 sin^2{\alpha} - 1}$ = 3 - 4 $\sin^2{\alpha}$ is equivalent to 16 $\sin^4$ $\alpha$ - 20 $\sin^{2}$ $\alpha$ + 5 = 0. (1). As a conclusion, the hypothesis of bisector theorem is verified, thus $\frac{BC}{AB}$ = $\frac{EC}{AE}$ => BE bisector, so m(BEC) = 18 => m(BCM) = 18 (because BE || MC) => m(ACM) = 36 + 18 = 54.