Quadrilateral ABCD has angles \(\angle{BAC}=80^{\circ}\) , \(\angle{CAD}=20^{\circ}\) , \(\angle{BDA}=50^{\circ}\) , and \(\angle{CDB}=50^{\circ}\) . Find angles \(\angle{BCA}, \angle{CBD}\)
Add a point E on line CD such that \(BE \parallel AD\)
It’s straightforward to establish that AB=DE, and that segment AE serves as the angle bisector, similar to segment BD. Consequently, we find that AC=CE, AB=BE=DE, then, \(BC \perp AE, \angle BCA=60^{\circ}, \angle CBD=10^{\circ}\)