Calculate the measure of the missing angle without trigonometry

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.

Solution 1:

Given: $\triangle ABC$ with $CE \perp AB$, $\angle ACE = 20^\circ$ and $\angle BCE = 10^\circ$. Also, there is a point $D$ on $CE$ extend such that $\angle CDA = 50^\circ$. The question asks us to find $\angle ABD$.

Here is a construction that works.

We choose a point $G$ on $CI$ extend such that $\angle ABG = 10^\circ$. Then $ACBG$ is cyclic and a quick angle chasing shows that $\angle CAG = \angle CBG = 90^\circ$. So $CG$ must be a diameter of the circumcircle of $ACBG$ and its midpoint $O$ is the circumcenter. That leads to $\triangle AOB$ being an equilateral triangle.

As $\angle ADC = 50^\circ$ and $\angle DCF = 10^\circ$, we see that $\angle AFB = \angle BAF = 40^\circ$. So $BF = AB = BO$.

Since $\angle OBC = 20^\circ$, we have $\angle OFB = \angle BOF = 10^\circ$

Then notice that $\triangle AOC \cong \triangle OBF$ (by S-A-S)

So it follows that $AC = OF = CH$ and since $\angle GOF = \angle GCH = 30^\circ$, $OFHC$ is a parallelogram and we obtain that $\angle BFH = \angle BCO = 20^\circ$.

$\angle AHD = \angle DAH = 40^\circ$. As segment $BD$ subtends $40^\circ$ at point $H$ and $F$ on the same side of the segment, $BDFH$ is cyclic. That leads to $\angle BDH = \angle BFH = 20^\circ$.

Finally, $\angle ABD = \angle AHD + \angle BDH = 60^\circ$.

Solution 2:

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 $\angle ABD=60^\circ$.

$\square$