The following is a problem I created that I am trying to solve:
Given quadrilateral $ABCD$ such that $AD = AC = AB = 10\sqrt{3}$ and $\angle DAC = 60^\circ$ and $\angle CAB < 90^\circ$, construct point $P$ on $BD$ such that $\overleftrightarrow{CP} \perp \overline{AB}.$ If $PB = 10,$ then what is $PD?$
By setting constructing a circle with center $A$, we have that points $B,C,D$ are all on the circle's circumference. which means that $\angle DBC = \angle DAC /2 = 30^{\circ}$.
After making this on geogebra, I got that $PD = 20$ and that ABCD should just be a rhombus, but I am having some trouble proving these.
A solution that should be straightforward is by applying coordinates: Letting $A$ be origin and $BC$ parallel to the x-axis. Then we can let $D=(a,b)$.
We can now find the equations of the BD and the perpendicular line and find their intersection, thus finding point $P$ in terms of $a$ and $b$.
We can then apply distance formula from $P$ and $B$ and equate it to $10$.
Our second equation comes from $a^2+b^2 = (10\sqrt{3})^2$ since $P$ on a circle with radius $10\sqrt{3}$.
However, this solution has really bashy calculations for such a simple configuration.
Any help would be greatly appreciated!