An Olympiad geometry problem ; angle chasing / trig bashing / coordinate bashing

Question

D is an inner point in equilateral triangle $ABC$ such that $\angle ADC=150^{\circ}$.Prove that the triangle formed by taking segments $AD$,$BD$ and $CD$ as its three sides is a right angled triangle. ...... [Source : Some European Olympiad in 2003]

This is the first time I am trying a problem like this (proving that these three lines make a right triangle).

I did manage to prove the special case when D $\in BM$ where M is midpoint of AC using simple sin rule, in which case I got $AD=CD=\frac{BD}{\sqrt2}=\frac{\sqrt3 - 1}{\sqrt 2}a$ where $a$ is the side length.

I did manage to find a solution online by AMAR_04 (Geo god on aops) (https://artofproblemsolving.com/community/c869463h1833925_build_up_your_understanding2_p21) but I am not very comfortable with rotation yet.

So, if anyone can please provide a different solution (trig bash/ coordinate bash or anything else except complex) OR explain the solution above, I would be really greatful!

Thanks!

Answer 1

Simple solution: by analytic geometry, the locus of points $P$ such that $PC^2=PA^2+PB^2$ is a circle.
This circle clearly goes through $A$ and $B$ and through some point $Q$ on the $AM$ median. Once proved that $\widehat{AQB}=150^\circ$ we are done. Without words:

Answer 2

Here is the figure for the solution on AOPS.

Rotate the whole figure through $B$ by $60^\circ$ anticlockwise.

By construction, $CE = AE'$, $AE = DE'$, $BE = BE'$, $\angle EBE' = 60^\circ$.

The last two conditions assures that $\triangle EBE'$ is equilateral.

Hence $\triangle AEE'$ is a right-angled triangle, formed by the side lengths $AE, BE, CE$.