geometry
For reference:In figure ABCD is a parallelogram and the triangles APB, AQD and CRD are equilateral. Find x. (answer: $120^\circ$)
My progress:
I found the following links and put it in the drawing
$AF = FC\\
BF = FD\\
\measuredangle QEG = 60^o\\
\measuredangle DAC = \measuredangle ACB$
but I can't finish..
$\triangle ABC$ is isosceles. Therefore,
$\measuredangle BAC = \measuredangle C \implies 2\alpha+\alpha =120^\circ\implies \alpha =40^\circ$
Therefore $\measuredangle ABC = 180^\circ - 160^\circ = 20^\circ$
Draw $AX$ such that $\measuredangle XAK = 30^\circ$.
Draw $TX$.
$\measuredangle CDB = 120^\circ ~~\therefore \measuredangle DIA = 90^\circ$
Since $CI$ is angle bisector $IA = IX$
$\triangle ITA \cong \triangle IXT ~(L.A.L) \rightarrow \measuredangle AXT = 60^\circ$
Therefore $\triangle ATX$ is equilateral.
$\measuredangle KXB = 180 ^\circ -110^\circ= 70^\circ$
$AK$ is angle bisector and the height of triangle $ATX$
$ \therefore AKT = 90^\circ$
$TK = KX \rightarrow \triangle KTB \cong \triangle KXB~ (L.A.L)$
Therefore $\triangle TBX$ is isosceles.
$\measuredangle B = 180^\circ -140^\circ =40^\circ$
$\therefore \boxed{\color{red}x=40^\circ-20^\circ=20^\circ}$
Let $K\in AD$ such that $ABCK$ be a parallelogram.
Thus, by law of sines for $\Delta KCD$ we obtain: $$\frac{m}{\sin\theta}=\frac{\frac{11m}{5}}{\sin3\theta}.$$ Can you end it now?
I got $\theta=\arcsin\frac{1}{\sqrt5}.$
Best Answer
Please note that $\angle PAQ = \angle BAD$, $AP = AB$, $AQ = AD$.
So $\triangle PAQ \cong \triangle BAD \ $ and hence $\angle AQP = \angle ADB$
Similarly show that $\triangle RDQ \cong \triangle CDA$ and so, $\angle DQR = \angle CAD$
But $\angle ADB + \angle CAD = 180^\circ - x \ $ (see $\triangle AOD$ where $O$ is the point of intersection of diagonals in the parallelogram $ABCD$)
That leads to,
$\angle PQR = x = 180^\circ - x + 60^\circ \implies x = 120^\circ$