In Quadrilateral $ABCD$, $AB=BC=AC$, $DC=3$, $AC=5$ and $\angle ADC=60$. Find the measure of $BD$ As title suggests, in the following figure with some given angles and sides, the goal is to find the measure of length $BD$. This problem originates from Iran from a math olympiad for high school students. I'll post my own approach below as an answer, please let me know if there any faults in it. And please post you own solutions too!

 A: Here's my solution

1.) First we'll notice that $\angle DAC+\angle DCA=120$. We can label $\angle DAC=\beta$ and $\angle DCA=\alpha$. We can "shift" $\triangle ADC$ over segment $AB$ such that the new triangle $\triangle AEB$ is congruent to $\triangle ADC$. Notice that since $\angle BAC=60$, $\angle EAB=\alpha$ and $\angle DAC=\beta$, we can conclude that segment $ED$ is a straight line, or in other words, point $E$, $A$ and $D$ are all collinear since $60+\alpha +\beta=60+120=180$. Note also that $\angle AEB=60$, $AE=3$ and $BE=5$.
2.) Now we know that $ED=8$, we draw a perpendicular from $D$ onto segment $BE$ at point $F$. We can immediately conclude that $\triangle DEF$ is a $30-60-90$ triangle. Therefore $EF=4$ and $DF=4\sqrt3$. We can also tell that $FB=1$. Finally, we can apply the Pythagorean theorem on $\triangle BDF$.
$$BD^2=FD^2+FB^2=48+1=49$$
$$BD=7$$
A: Here is an alternative solution for a problem formulated in a slightly more general setting. (We use $x,y$ instead of $3,5$.)

Problem: Let $\Delta ABC$ be equilateral. Construct $B'$, the reflection of $B$ w.r.t. $AC$. (So $B'\ne B$ is characterized by making  $\Delta AB'C$ also equilateral.) Consider a point $D$ on the circumcircle of $\Delta AB'C$ with $\widehat{ADC}=60^\circ$. Assume that the segments $CD$ and $AD$ have lengths $x$ and $y$ respectively.
Then: $BD^2=x^2+y^2+xy$.

Proof:
Consider the rotation $r$ with center $B$ and angle $60^\circ$ that brings $C$ to $C^r=A$. Denote by $D^r$ the image of $D$. The rotation of $\Delta BCD$ is $\Delta BAD^r$, so $AD^r=x$, and since we rotate with an angle of $60^\circ$ the triangle $BDD^r$ is equilateral. (The following picture is already the proof.)

We move our focus on $\Delta DAD^r$ and want to show that its $\hat A$ is $120^\circ$. This is so because:
$$
\widehat{DAD^r}
=
360^\circ -
(
\widehat{D^rAB}+
\widehat{BAD}
)
=
360^\circ -
(
\widehat{DCB}+
\widehat{BAD}
)
=
\widehat{ABC}+
\widehat{ADC}
=60^\circ+60^\circ\ .
$$
Now apply the generalized version of the theorem of Pythagoras in $\Delta D^rAD$, and use $\cos \hat A=\cos 120^\circ =-1/2$ to conclude:
$$
BD^2 =D^rD^2=x^2+y^2-2xy\cdot\left(-\frac 12\right)
=x^2+xy+y^2\ .
$$
$\square$

In the given problem, we compute $x^2+xy+y^2=3^2+3\cdot 5+5^2=49=7^2$, and take the square root to obtain $BD$.
