D is a point on side AC with AD=BC. Find angle DBC.

Solve this without trigonometry.

Find the point E such that EA = ED = AB, with E and B both on the same side of line ADC.

Then: triangles AED and BAC are congruent triangle EAB is isosceles angle AED = 20 angle EAD = 80 angle EAB = 60 triangle EAB is equilateral angle AEB = 60 angle DEB = 40 EA=EB=ED triangle DEB is isosceles angle DBE = angle BDE = 70 angle EBA = 60 angle EBC = 140 angle DBC = 70 degrees

blackjack

flooble's webmaster puzzle