Two equal segments, two known angles, find angle in triangle

Question

This is the problem. I have tried the obvious without success. Drawing some perpendiculars (from A for example), I only know the angle $\sphericalangle ADC = 54 ^{\circ}$. How can I use the equality $AB = DC$?

Answer 1

By the law of sines we obtain: $$\frac{AD}{\sin{x}}=\frac{DC}{\sin(126^{\circ}-x)}$$ and $$\frac{AD}{\sin24^{\circ}}=\frac{AB}{\sin54^{\circ}}$$ and since $AB=DC$, we obtain: $$\frac{\sin{x}}{\sin24^{\circ}}=\frac{\sin(126^{\circ}-x)}{\sin54^{\circ}}.$$ Can you end it now?

I got $x=30^{\circ}.$

Answer 2

Let's reformulate the problem.

Let $\triangle ABC$ be such that $\angle ABC = 24^{\circ}$ and $\angle ACB = 30^{\circ}$ and $D$ is on $BC$ so that $AB = CD$.

We have to prove $\angle BAD = 30^{\circ}$

Solution: Rotate $B$ around $A$ for $60^{\circ}$ to new point $E$. Then since $EA=EB$ and $$\angle AEB = 2\angle BCA$$ we see that $C$ is on circle with center at $E$ and radius $EA=EB = EC$. Then $CD = CE$, so $$\angle BCE = \angle CBE = 34^{\circ} \implies \angle EDC = \angle DEC = 72^{\circ}$$ Now this means that $$\angle BDE = 108^{\circ} \implies \angle BED = \angle EBD = 36^{\circ}$$ This means $BD = ED$, so so $\triangle ABD$ and $\triangle ADE$ are congruent (sss) and thus $\angle BAD = 30 ^{\circ}$.