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 $
On
I was searching for a geometric solution inside a "very symmetric setting", but soon after posting a solution was accepted, so i gave up mainly for time reasons, i had no "easy" argument on my way at that point. However, after some weeks i came by chance across the older notices, and could complete the proof with an elementary bridge, so i decided to share the ideas. Problems like the given one can almost always be connected with special configurations of points and lines related to some regular polygon. To give an example, the well known Langley problem deals with configurations of lines building angles which are multiples of $10^\circ$. The wiki page on the Langley's Adventitious Angles mentions that generalizations of the Langley configuration is related to regular polygons. Citing from the wiki page:
Classifying the adventitious quadrangles (...) turns out to be equivalent to classifying all triple intersections of diagonals in regular polygons.
In our case the angles (like $18^\circ$, $24^\circ$, $30^\circ$) appearing are multiples of $6^\circ$, so one can try to find a solution related to a $30$-gon. Here is such a possible solution in the spirit of finding geometric properties of the regular $30$-gon equivalent to the "adventitious (quadr)angles" from the stated problem.
Let us realize the given triangle $\Delta ABC$ as the triangle $$\Delta 707'$$ inside the regular $30$-gon $\Pi$ with vertices labelled $0,1,2,3,\dots,29\in\Bbb Z/30$, the figure is below. The labels are integers considered modulo $30$. It is convenient to use a prime instead of a minus for the reflection w.r.t. the line $0\;15$, so $-X= X'$, making the notation of triangles, angles, etc. less operational. So for instance $7'=-7=30-7=23$ are equivalent labels for the point $C$. Let $*$ be the reflection w.r.t. the side bisector of $BC$. Reflect $BD$ by $*$ to get the segment $B^*D^*=CD^*$ of length $BD=AC=77'$.
Then the triangle $\Delta ACD^*=\Delta 77'D^*$ is isosceles in $C=7'$, and since $D\in 73'$, we have $D^*\in (73')^*=7^*{3'}^*=14'4'$. The needed angle $\widehat{ADC}$ is reflected in $\widehat{14'D^*0}$, so it is an equivalent task to compute this angle. There is an advantage of involving $D^*$ instead of $D$, because we have already an isosceles triangle $\Delta CD^*A$ to work with, and because of the following claim of concurrence of lines related to the $30$-gon. (The point $D^*$ is the point $X$ in the Claim below.)
(The figure is heavily decorated, as it often happens when searching for a solution, please extract only the used ingredients from it, or use the other visible coincidences to find an other solution.)
It is enough to show the concurrence of the lines $07=BA$, and $4'14'$, and $7'9'=C9'$ in order to solve the given problem, because knowing this we have: $$ \widehat{ADC} = \widehat{AD^*C} = \text{Angle between $07$ and $4'14'$} \\ =\frac12 \left( \overset\frown{14'7} - \overset\frown{4'0} \right) =(9-4)\cdot 6^\circ =30^\circ\ . $$ But let us show more:
Claim: In the given $30$-gon $\Pi$ consider the perpendicular bisector of the diameter $5\;10'$ as an axis of symmetry for a reflection $r$. Then the following pairs of $r$-reflected lines are concurrent in a point $X$ of this axis: $$ \begin{aligned} 24 &\text{ and } 7'9' \ , \\ 07 &\text{ and } 5'12' \ , \\ 1'9 &\text{ and }4'14' \ . \end{aligned} $$
I need a Lemma first:
Lemma: In the given $30$-gon $\Pi$ the diameter $78'$, and the pair of reflected lines w.r.t. to this diameter
- $7'13$ and $9'1$,
- $4'11$ and $5'12$
are passing through a point $H$.
Proof of the Lemma: Define $H\in 78'$ as the intersection $H=9'1\cap 7'13$. Then $$ \Delta 13H9'\sim \Delta 75'7' \ , \\ \qquad\text{ since }\hat{13}=12^\circ=\hat 7\text{ and } \hat 9'=(13-1)\cdot 6^\circ=(7-(-5))\cdot 6^\circ = \hat 7'\ . $$ The position of $H$ on $7'13$ can be computed as: $$ \frac{H7'}{H13} = \frac{H9'}{H13} = \frac{5'7'}{5'7} = \frac{5'7'}{5'13} \ . $$ Then in $\Delta 5'7'13$ the line $5'H$ is an angle bisector, showing that $5',H,12'$ are colinear.
$\square$
Proof of the Claim: Consider the triangle $\Delta 799'$, and some of its internal an external angle bisectors.
From here, the $9$-excenter of $\Delta 799'$ is the intersection of the internal, and respectively external angle bisectors $1'9$, and respectively $07$, $7'9'$. Let us denote this point by $X$.
(The other three, $r$-reflected lines from the claim are also concurrent in the point $X^r$. Let us show $X=X^r$.)
Let us consider the heights in the triangle $\Delta X79'$:
The height from $7$ is $78'$ since $78'\perp 7'9'$ and $X$ is on this last line. The height from $9'$ is $9'1$ since $9'1\perp 07$ and $X$ is on this last line. From the Lemma, these two heights intersect in a point $H$ which is also on $5'12'$, and we note that $5'12'$ is perpendicular on the side $79'$ of $\Delta X79'$. So the height from $X$ is this line $5'12'$.
This shows $X\in 5'12'$. But $X$ is already also on $r$-reflection of $5'12'$, which is $1'9$. S $X=X^r$. This completes the proof of the Claim.
$\square$
On
Equation AB = BC implies that triangle ABC is isosceles
Moreover we have given measure of one angle so we can calculate measure of other angles
and make measure of angle BCD dependent on x
Equation AB = BC implies that triangle ABC is isosceles then base angles have equal measure
From sum of angles measures in ABC we get that measures of angles are equal 42 , 42, 96 in degrees
From sum of angles measures in ACD we get
$$
24+42+y+x = 180\\
66+x+y = 180\\
x+y=114\\
y=114-x\\
$$
Now we want to make dependent lengths of the sides in triangle BCD on one side length only
We use law of sines twice to do this
From law of sines in triangle ABC
$$
\frac{AC}{\sin(96)}=\frac{BC}{\sin(42)}\\
\frac{AC}{\sin(84)}=\frac{BC}{\sin(42)}\\
\frac{AC}{2\sin(42)\cos(42)}=\frac{BC}{\sin(42)}\\
BC = \frac{AC}{2\cos(42)}\\
BC = \frac{BD}{2\cos(42)}\\
$$
From law of sines in triangle ACD
$$
\frac{AC}{\sin(x)} = \frac{CD}{\sin(24)}\\
CD = \frac{\sin(24)}{\sin(x)}AC\\
CD = \frac{\sin(24)}{\sin(x)}BD\\
$$
From law of cosines in triangle BDC
$$
BD^2=\frac{BD^2}{4\cos^2(42)}+\frac{\sin^2(24)}{\sin^2(x)}-2\frac{\sin(24)}{\sin(x)}BD\frac{BD}{2\cos(42)}\cos(114-x)\\
BD^2=BD^2\left(\frac{1}{4\cos^2(42)}+\frac{\sin^2(24)}{\sin^2(x)}-\frac{\sin(24)}{\sin(x)\cos(42)}\cos(114-x)\right)\\
1 = \frac{1}{4\cos^2(42)}+\frac{\sin^2(24)}{\sin^2(x)}-\frac{\sin(24)}{\sin(x)\cos(42)}\cos(114-x)\\
$$
This equation can be expressed as quadratic in terms of $\cot(x)$
$$ 1 = \frac{1}{4\cos^2(42)}+\frac{\sin^2(24)}{\sin^2(x)}-\frac{\sin(24)}{\sin(x)\cos(42)}\cos(114-x)\\ 1 = \frac{1}{4\cos^2(42)}+\sin^2(24)\frac{\cos^2(x)+\sin^2(x)}{\sin^2(x)}+\frac{\sin(24)}{\cos(42)}\frac{\cos(66)\cos(x)-\sin(66)\sin(x)}{\sin(x)}\\ 1 = \frac{1}{4\cos^2(42)}+\sin^2(24)\frac{\cos^2(x)+\sin^2(x)}{\sin^2(x)}+\frac{\sin(24)}{\cos(42)}\frac{\sin(24)\cos(x)-\cos(24)\sin(x)}{\sin(x)}\\ 1 = \frac{1}{4\cos^2(42)}+\sin^2(24)\left(\cot^2(x)+1\right)+\frac{\sin^2(24)}{\cos(42)}\cot(x)-\frac{\sin(24)\cos(24)}{\cos(42)}\\ 1 = \frac{1}{4\cos^2(42)}+\sin^2(24)\left(\cot^2(x)+1\right)+\frac{\sin^2(24)}{\cos(42)}\cot(x)-\frac{2\sin(24)\cos(24)}{2\cos(42)}\\ \frac{1}{4\cos^2(42)}+\sin^2(24)\cot^2(x) + \sin^2(24) - 1 + \frac{\sin^2(24)}{\cos(42)}\cot(x)-\frac{\sin(48)}{2\cos(42)}=0\\ \sin^2(24)\cot^2(x)+\frac{\sin^2(24)}{\cos(42)}\cot(x)+\frac{1}{4\cos^2(42)}-\cos^2(24)-\frac{1\cos(42)}{2\cos(42)}=0\\ \sin^2(24)\cot^2(x)+\frac{\sin^2(24)}{\cos(42)}\cot(x)+\frac{1-4\cos^2(42)\cos^2(24)-2\cos^2(42)}{4\cos^2(42)} = 0\\ \cot^2(x)+\frac{1}{\cos(42)}\cot(x)+\frac{1-4\cos^2(42)\cos^2(24)-2\cos^2(42)}{4\cos^2(42)\sin^2(24)}=0\\ \left(\cot(x)+\frac{1}{2\cos(42)}\right)^2-\frac{1}{4\cos^2(42)}+\frac{1-4\cos^2(42)\cos^2(24)-2\cos^2(42)}{4\cos^2(42)\sin^2(24)}=0\\ \left(\cot(x)+\frac{1}{2\cos(42)}\right)^2-\frac{1}{4\cos^2(42)}-\frac{4\cos^2(42)\cos^2(24)+2\cos^2(42)-1}{4\cos^2(42)\sin^2(24)}=0\\ \left(\cot(x)+\frac{1}{2\cos(42)}\right)^2-\frac{4\cos^2(42)\cos^2(24)+2\cos^2(42)-1+\sin^2(24)}{4\cos^2(42)\sin^2(24)}=0\\ \left(\cot(x)+\frac{1}{2\cos(42)}\right)^2-\frac{4\cos^2(42)\cos^2(24)+2\cos^2(42)-\cos^2(24)}{4\cos^2(42)\sin^2(24)}=0\\ $$
$$
\cot(x)=\frac{-\sin(24)\mp\sqrt{4\cos^2(42)\cos^2(24)+2\cos^2(42)-\cos^2(24)}}{2\cos(42)\sin(24)}\\
$$
but only root with + sign works
It should be simplified further but value of RHS is $\cot(162)$ and $\cot(30)$
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 $\angle ADC = \angle CDF - \angle ADF = 30^\circ$.