As the title says, in the setup given below, we have an isosceles triangle $\triangle ABC$ where the two base angles are $30^\circ$ each. $P$ and $Q$ are located on $AC$ such that $AP=2$ and $QC=1$, and $\angle PBQ=60^\circ$. The goal is to find the measure of length $PQ$:
This problem was posted by a user on a languages learning app that I came across. The claim is that it is meant for 10th grade students. The user did not share the correct answer.
At first I attempted to "complete the triangle" by forming a large equilateral triangle, but that approach didn't lead anywhere. I will share my only successful approach below as an answer. Please share your own answers as well and let me know if the answer I arrived at is correct.
Best Answer
This is my approach:
1.) Draw a line from $B$ onto point $D$ outside of $\triangle ABC$ such that $\angle ABD =\angle CBQ$ and $BD=BQ$. Connect $A$ and $D$ via $AD$ and notice that $\triangle ABD$ is congruent to $\triangle CBQ$. Also join $D$ and $P$ via $DP$ and $D$ and $Q$ via $DQ$. Notice that $\angle DAP=60^\circ$ and that, $AP=2$ and $AD=CQ=1$, this means that $\triangle DAP$ is a "$30-60-90$" triangle.
2.) Above implies that $DP=\sqrt{3}$ and $\angle APD=30^\circ$. Now notice that $\angle ABP+\angle ABD=60^\circ$, this means that line segment $BP$ is the perpendicular bisector of the isosceles triangle $\triangle DBQ$. This means that $DE=EQ$. But, this also implies that $\triangle DPQ$ is an isosceles triangle. Therefore, $PQ=DP=\sqrt{3}$