Geometria (solo trazos), reposted, https://www.facebook.com/photo?fbid=542149486954170&set=gm.1141154929709218

 In convex quadrilateral ABCD <BAC=48, <CAD=18, <ADB=30, <ACB=6, find <ACD.

Proof with Mustafa Yagci triangle
Take O the circumcenter of triangle ABC; as <ABD=90-<ACB, O lies onto BD and <AOD=<AOB=2<ACB=12 ( 1 ). Construct the equilateral triangle ACE, D inside it; as O lies onto the perpendicular bisector of AC, it follows that <AEO=30=<ADB and ADOE is cyclic, wherefrom <AED=<AOD=12. With <DAC=18 and (1) we get from Mustafa Yagci triangle <ACD=6, done.