Angle Chasing in a convex quadrilateral

Question

Find the measures of the angles of the convex quadrilateral ABCD, if $\angle ACD = 78°$ , $\angle BDC = 22°$, $\angle CBD = 12°$ and $\angle CAD = 24°$.

Source: Romanian Mathematical Gazette

Answer 1

It seems that in the quadrilateral $AB=AC=AD$.

Thus $\triangle ACD$ is isosceles and$$\angle ADC=78^o$$Next, since $\triangle ABD$ is isosceles, then$$\angle ABD=\angle ADB=78-22=56^o$$so that $$\angle ABC=56+12=68^o$$Again, since $\triangle ABC$ is isosceles, then $\angle ACB=\angle ABC=68^o$, and$$\angle BCD=68+78=146^o$$Lastly, since $\angle CAB=180-2\cdot 68=44^o$, then$$\angle DAB=44+24=68^o$$

Answer 2

After calculating some angles with the given information we find that $\angle ADC = \angle ACD = 78^\circ$. Therefore $\triangle ACD$ is isosceles. Let $P$ and $P'$ be the intersection of the bisector of angle $A$ with diagonal $BD$ and $CD$. Since $\triangle ACD$ is isosceles and $AP'$ is the angle bisector of $A$ it results that $AP'$ is also perpendicular on $CD$ and $P'$ is the midpoint of $CD$. Thus $PP'$ is the median and height in the $\triangle DPC$.

Hence it follows that $\triangle DPC$ is isosceles. Then $\angle PDC = \angle PCD = 22^\circ$. Thus $\angle ACP =56^\circ$. Since $ABCP$ is cyclic $\angle ABP = \angle ACP = 56^\circ$. Hence $\angle BAC = 100^\circ-56^\circ = 44^\circ$.