TITLE: Calculate the measure of the missing angle without trigonometry
QUESTION [2 upvotes]: This was from a question that was closed recently.
It says to find the measure of angle '$a$'. (answer: $60^\circ$)

The below is my attempt to solve the question.

As you'll see I drew $\triangle DAC$ that is congruent to $\triangle CFD$.
And I drew $\triangle CGD$ congruent to $\triangle DBC$.
Also, I drew parallelogram $FGDI$.
But any of these weren't useful to solve find the measure of angle $a$.
I can find the value using trigonometry.
But sometimes trigonometry becomes boring so anyone in this community could help me to solve this question without using trigonometry?
Thank you.

REPLY [1 votes]: Here is a much simpler solution.
Mirror $B$ across $AC$.

It is easy to verify the collinearity of points $B'$, $A$, $D$. Then notice that $\triangle CBB'$ is equilateral and $\triangle CB'D$ is isosceles. Therefore $CB'=BB'=DB'$ which implies $B'$ being the circumcentre of $\triangle CBD$. That gives $\angle CDB=\frac12\angle CB'B=30^\circ$ and hence $\angle ABD=60^\circ$.
$\square$
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