Tricky Triangle Analysis

We will prove $AC = AD$ , which will allow to easily obtain the requested angle.

By interior angle sum in triangle $ABC$ we easily get $\angle CAB = \angle CBA = 42^\circ$ . Construct equilateral triangle $ABE$ such that $C$ lies inside it. In order to prove $AC = AD$ we will prove triangles $ACE$ and $ADB$ are congruent.

Since $\angle BAE = 60^\circ$ we get $\angle CAE = \angle EAB - \angle CAB = 60^\circ - 42^\circ = 18^\circ$ .

We know $CA = CB$ , which imples $C$ lies on the perpendicular bisector of $AB$ , therefore, since triangle $ABE$ is equilateral, we have $\angle CEA = \angle CEB = 30^\circ$ .

Therefore, since $\angle CAE = \angle DAB = 18^\circ$ , $\angle AEC = \angle ABD = 30^\circ$ and $AB = AE$ , we get that triangles $ABD$ and $AEC$ are congruent by ASA, which implies:

$AC = AD$

Therefore $\angle ACD = \angle ADC$ , and since $\angle CAD = \angle CAB - \angle DAB = 42^\circ - 18^\circ = 24^\circ$ , by interior angle sum in triangle $ACD$ we get

$\angle ACD = \frac{180^\circ - 24^\circ}{2} = 78^\circ$ .

Diagram

By angle sum property of triangle one will find that $angle$ $CAD$ is $24°$ and $angle$ $ADB$ is $132°$ . Let angle $ACD$ be $x$ . Then angle $ADC$ will be $156-x$ . Let's assume that $AB=1$ . Then by applying law of sines for triangle ABC we have $AC/sin 42 = 1/sin 96$ . $AC= sin 42/sin96$ . Again by applying law of sines for traingle ABD we find that $AD=1/2sin48$ . After using law of sines in triangle ACD we have the following equation: $1/(2sin48*sinx)=sin42/(sin96*sin (156-x))$ . After simplification we are left with $sin(156-x)=sinx$ . By comparison $x=156-x$ or $x=78°$

CAD is 24º. Them draw the bissector CH of angle ACB and let AD intersect CB in N. Draw the bissector AS of angle CAN and Let it intersect CH in K. Notice triangle AK = KB, angle HKB = BKS = SKC = 60º; Angles AKC = AKD = 120º then triangles ACK and ADK are congruents by SAS, and because it, CK = KD. Notice KS is bissector of triangle isosceles CKD. Let M be the midpoint of CD, and we have angle KMC = 90º, and angle KCM = 30º. As angle ACD = ACK + KCM, we have angle ACD = 48º + 30º = 78º.

We claim that $\angle ACD=\boxed{78^{\circ}}$ .

To prove this, let $D'$ be the point in triangle $ABC$ such that $\angle D'AB=18^{\circ}$ and $\angle D'CA=78^{\circ}$ . This implies $\angle D'AC=24^{\circ}$ and $\angle AD'C=78^{\circ}$ , so triangle $ACD'$ is isosceles with $AD'=AC$ . Now construct $E$ such that $CAD'E$ is a parallelogram. Since $AD'=AC$ , it is in fact a rhombus. Furthermore $\angle ECB=180^{\circ}-96^{\circ}-24^{\circ}=60^{\circ}$ and $EC=BC$ , so triangle $BCE$ is in fact equilateral.

Next, note that $\angle D'EC=\angle D'AC=24^{\circ}$ , so $\angle D'EB=60^{\circ}-24^{\circ}=36^{\circ}$ . Since triangle $D'EB$ is isosceles, it follows that $EBD'=72^{\circ}$ . Finally, $\angle D'BC=72^{\circ}-60^{\circ}=12^{\circ}$ which means $\angle D'BA=30^{\circ}$ . Since there is a unique point $P$ in triangle $ABC$ satisfying $\angle PAB=18^{\circ}$ and $\angle PBA=30^{\circ}$ , we conclude that $D'=D$ , so $\angle ACD=\angle ACD'=78^{\circ}$ .

Or we could just solve this problem using Trig Ceva.

$Because~ \Delta~ABC~ is~ isosceles~ with~ vertex ~\angle~96^o, ~\angle~ DAC=24^o~ and ~\angle~ DBC=12^o.\\ Also ~\angle~ BDA=180-18-30=132^o.~ If~ \angle~ CDA=X,~ angle~ CDB=228 - X.\\ Applying ~Sin~ Law ~to~ \Delta s~ CAD ~and~CDB,\\ \dfrac{SinCAD}{SinADC}=\dfrac{CD}{CA}=\dfrac{CD}{CB}=\dfrac{SinDBC}{SinCDB}.\\ \implies~\dfrac{Sin24}{Sin(228-X)} = \dfrac{Sin12}{SinX}.\\ Expanding~ Sin(228-X)~ and~ Sin2A=2SinA*cosA,~ on ~simplifying,\\ CotX=\dfrac{\frac 1 {2cos12} +Cos228}{Sin228}.~\implies~X=78^o.\\ \therefore~\angle~ACD=360-24-78=\Large~~\color{#D61F06}{78^o}$

assume acd to be x, angle cad is 24,side ad is given by applying sine rule in triangle adb,given by a sin30/sin132,where a=ab.use sine rule for side ad and ac ,ac is given by a sec42/2,sine rule in triangle acd simplifies into sin(156-x)=sinx. 156-x=x x=78