euclidean-geometrygeometryplane-geometry
In the figure, A, B, C, and D are points of tangency . Determine x. (S:$x=60^o$)
I try
$\angle FUG = x+3(\beta+\theta)$
$\angle CFE = 180-4\theta-x (I)$
$\angle BGE = 180-4\beta – x(II)$
$(I)and(II): \angle CFE -\angle BGE = 4(\beta-\theta)$
$\angle BFC \cong \angle UGC$
Add another congruent triangle $BCD$ and put point $E$ in it such that $DE=CP, ~ BE=BP, ~ CE=AE.$ Then $\angle PBE = 90°,$ $PE = 2x\sqrt{2}$. Then $\angle CPE = 90°.$ Therefore $\angle BPC = 135°.$
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
I will use $t,s$ instead of $\theta,\beta$ for an easy typing. We are starting with the following situation, which copies the OP figure, and marks some more angles relevant for a first step:
Let $u$ be the common measure of the arcs $ \overset\frown{AB} = \overset\frown{BC} = \overset\frown{CD} $. Then the complement of $u$, i.e. $90^\circ-u$, as built in $F$ and in $G$ gives: $$ \frac{4s+x}2=u=\frac{4t+x}2\ ,\ $$ which implies $s=t$. (Above, $4s+x$ is the exterior angle in $G$ for$\Delta BEG$, so it is the sum of the interior angles in $B,E$, which were given.) From now on we will keep only $s$, use it also instead of $t$. The sum of the angles in the quadrilateral $OAED$ is $360$, which gives also $x$ in terms of $s$ from $180^\circ=3u+(t+x+s)=3\left( 90^\circ - \frac{4s+x}2 \right)+(2s+x)$, we extract first $4s+x$, then $x$: $$ x = 180^\circ -8s\ . $$ Then we can compute other angles in terms of $s$: By symmetry, $U$ is on $EO$, and $EO\perp FG$. The angle $\widehat{FEU}$ is $\frac 12(t+x+s)=90^\circ-3s$, so $\widehat{EFG}=\widehat{EGF}=3s=\widehat{EBG}$, the quadrilateral $EFBG$ is cyclic, so $EFBCG$ cyclic.
Before we conclude, it is maybe a good didactic idea to show the situation so far. For a given $s$ we have then the following picture, some of the properties of the given picture have been used, but evidently, some of them are missing:
(A lot of things do not fit, $EFL$, $EGM$ are not tangent $FUC$, $GUB$ are not tangent to the circle $O$. How to conclude quickly?)
From $EFBCG$ cyclic we obtain $\widehat{FBE}=\widehat{FCE}=3s$. Recall that we want a right angle $$ 90^\circ=\widehat{FBG}=\widehat{FBE}+\widehat{EBG}=3s+3s=6s \ . $$ This determines $s=15^\circ$, and then $x=180^\circ-8s=60^\circ$.
$\square$
Note: So far we have shown that if such a configuration exists, then $s=15^\circ$ and $x=60^\circ$. Well, the problem assumed that it exists, and we may stop here. But let us prove that the only one possible configuration exists indeed. We do so from the second figure. If $s=15^\circ$, then conversely $\widehat{FBG}=3s+3s=90^\circ$, so $BG$ is tangent to the circle $(O)$ in $B$. Why is $EFL$ tangent to this circle, too?
We compute angles of $\color{forestgreen}{30^\circ}$ in $L,U,M$ as shown in the picture. Let $O$ be the intersections of the lines $FB$ and $GC$. These are the perpendicular bisectors of the segments $LU$, $UM$. The reflection of $FUC$ w.r.t. $FBO$ is $FL$ first, and it is also tangent (in the reflection of $C$) to the circle. In same manner, $GM$ is a tangent, too.