euclidean-geometrygeometryplane-geometry
For reference: If $AB = BC$ and $AC = BD$ find x(Answer:$30^o$)
I found the possible Angles, I drew some auxiliary lines ($DG \parallel AC, NH \perp AC$) but I still couldn't find the relationship for the solution
$\alpha+x+84-\beta +\theta = 180 \implies x = 96-\alpha+\beta-\theta\\
\beta +24-\alpha = 180-138 \implies \beta-\alpha = 18\\
x =114 – \theta $
Build a parallelogram $ABDE$. Then $$\angle DEA = \angle ABD = 135^\circ = \angle DCA,$$ hence $ADEC$ is cyclic. Moreover, $\angle EAD = \angle BDA = x$, hence $$\angle CAE = \angle CAD - \angle EAD = 2x-x=x.$$ As a consequence, $CE=DE$ because the angles subtended by arcs $CE$, $DE$ of the red circle are equal.
Now, it is easy to calculate that angle $CBD$ equals $90^\circ + x$ and that the concave angle $CED$ equals $180^\circ + 2x$. Since $CE=DE$, it follows that $B$ lies on the circle with center $E$ and radius $CE=DE$. Hence $\angle DEB = 2\angle DCB = 90^\circ$. Since $DE=BE$, the right triangle $DEB$ is isosceles.
Let $F$ be the midpoint of $BE$. Let $G$ be the midpoint of $BD$ and $H$ be the midpoint of $BG$. Easy to see that triangles $GFB$ and $GFH$ are isosceles right triangles, hence $HF = HB = \frac 12BG = \frac 14 BD$, so $HD = BD - BH = 4HF - HF = 3HF$. This yields $\tan x = \frac{HF}{HD} = \frac 13$ so the answer is $$x = \arctan \frac 13.$$
If three angle bisectors of a quadrilateral are concurrent, the fourth angle bisector is concurrent with them (see angle bisectors section wiki). Drawing perpendicular segments to sides of quadrilateral from the point of concurrency of three angle bisectors easily shows it.
Now consider quadrilateral $ACEF$. As three angle bisectors meet at $I$, the fourth will too. In other words, $AI$ is angle bisector of $\angle BAC$.
So in $\triangle AHF$, $J$ is the incenter,
$\angle AHJ = 90^\circ - \theta - (50^\circ - \beta) ~$ (as $~\angle BAC = 100^\circ - 2 \beta$)
In $\triangle CEG$, $K$ is the incenter,
$\angle CGK = 90^\circ - \alpha - \beta$
So, $x = 180^\circ - (130^\circ - \theta - \alpha) = 115^\circ ~$ (as $~\theta + \alpha = 65^\circ$)
Best Answer
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 $\angle ADC = \angle CDF - \angle ADF = 30^\circ$.