Find angles of triangle formed by images of vertices about opposite sides of an isosceles triangle

Question

In $\triangle ABC$, $AB=AC$ and $\measuredangle BAC=30^\circ$. If $A^\prime$, $B^\prime$ and $C^\prime$ are the reflections of $A$, $B$, and $C$ about $BC$, $CA$ and $AB$, How to find $\measuredangle A^\prime B^\prime C^\prime$?

I tried to solve this using co-ordinate geometry (taking $\measuredangle BAC$ as a general $\theta$) and obtained a complicated relation $\tan\measuredangle B'A'C'$ = $\frac {2(5a^2b+b^3)(3ab^2-a^3)}{(3ab^2-a^3)^2+(5a^2b+b^3)^2}$ where $AB = AC = 2a$ and height $AD =b$, i.e. $\tan (\frac{\measuredangle ABC}{2}) =\tan (\frac{\theta }{2}) = \frac{a}{b}$.

I am not sure I am heading the right way.

How to do this using plain geometry?

Answer 1

First of all notice that $BC'=BC$, $\angle BC'B'=30°$ and $\angle A'BC'=135°$. These are very easy to prove, so I won't go into details. Let now $\angle BC'A'=\alpha$, so that $\angle BA'C'=45°-\alpha$, and let $BH$ be the altitude of triangle $A'BC'$ with respect to base $A'C'$. Then we have: $$ BH=BC'\sin\alpha=BC\sin\alpha\quad \hbox{and}\quad BH=BA'\sin(45°-\alpha)={BC/2\over\sin15°}\sin(45°-\alpha). $$ Equating these two expressions for $BH$ we get $$ 2\sin15°\sin\alpha=\sin(45°-\alpha). $$ Inserting here $\sin15°={\sqrt3-1\over2\sqrt2}$, the above equation reduces to $\sqrt3\sin\alpha=\cos\alpha$, that is $\alpha=30°$. It follows that $\angle B'C'A'=\angle BC'B'+\alpha=60°$.

Answer 2

Notice that $ABA'C$ is a parallelogram.
$\measuredangle B'BA'=\measuredangle BXC=90^{\circ}$

$BB'=2BX=2a \sin75^{\circ}$
$BA'=BA=\frac{a}{2\cos75^{\circ}}$

Observation: $2\sin75^{\circ}=\frac{1}{2\cos75^{\circ}}$
Implying $BB'=BA'$

You can continue from here.( You should get $\measuredangle CA'B'=15^{\circ}$ )

A pure geometry approach would be:
Observation: Triangle $ABB'$ is equilateral.That gives you $BB'=BA'$