euclidean-geometrygeometryplane-geometry
In the plane figure below, we have a regular octagon $ABCDEFGH$, and I belongs to the diagonal $CG$, so that $\angle GIH = 30°$. Knowing this, determine in degrees the value of the angle indicated by x.(S:$75^o$)
Itry:
$a_i=\frac{180.(8-2)}{8} = 135^o $
$\angle IGH = \frac{135} {2}=67,5^o \implies\angle IHG = 82,5^o$
$\angle AHB = \frac{180-135}{2}=22,5^o \implies \angle BHI =135 – 82,5-22,5 = 30^o \cong \angle GIH$
I'm missing a relationship to finish that I didn't find
Reflect point $A$ about $BC$. Then by midpoint theorem, $BM \parallel A'E$
Now note that $\triangle A'CE \sim \triangle BCD $ and so we have $\angle CA'E = \angle CBD$
Also, $\angle ABM = \angle AA'E = 45^\circ - \angle CA'E = 45^\circ - \angle CBD$
That shows $\angle DBM = 45^\circ$
As a side note, if you also reflect point $E$ about $CD$, you can easily show $AE' = A'E$ and that means $BM = DM$. So, $\triangle DMB$ is indeed an isosceles triangle, right at $M$.
Trigonometric solution: this can be solved using the identity $~2 \sin 48^\circ \sin 12^\circ = \sin 18^\circ$.
Using law of sines, $~ \displaystyle \frac{AB}{\sin \angle ADB} = \frac{BD}{\sin 18^\circ} $
Now we know that $BD = AC = 2 AB \sin 48^\circ$
So, $2 \sin 48^\circ \sin ADB = \sin 18^\circ \implies \angle ADB = 12^\circ$
From here, you can see that $F$ is on perp bisector of $AC$ such that $BD = FD = AF = FC$ and we obtain $\angle ADC = \angle CDF - \angle ADF = 30^\circ$. Alternatively, we can also use law of sines in $\triangle BCD$.
Geometric solution:
We first draw an equilateral triangle $\triangle AFC$. Then I take a point $D'$ on the shown line, such that $AF = FD'$. Now we extend $AF$ to $E$ such that $AE = AD'$. Then $D'E = FD' = AF = AC$. Now I find a point $B'$ on the perp bisector of $AC$ such that $B'D' = FD'$.
That means $~\angle B'D'F = 24^\circ$ and hence $\angle B'D'E = 60^\circ$ with $B'D' = D'E$. It follows that $~ \triangle B'D'E$ is equilateral and then $AB'$ must be angle bisector of $\angle D'AE$.
That shows points $B'$ and $D'$ used in our set up are the same points $B$ and $D$ as given in the question.
Using $FD = FC$ and $\angle CFD = 48^\circ$, we get $\angle ADC = \angle CDF - \angle ADF = 30^\circ$.
Best Answer
Draw $IF$ and $HF$ and by symmetry you have an equilateral triangle so $HI=HF=HB$ so you have an isosceles $\triangle {BHI}$ with 30 degrees in the apex and thus $75$ degrees in the base angles.