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!


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 $\triangle BDF$.
$$BD^2=FD^2+FB^2=48+1=49$$
$$BD=7$$
