A geometry problem by A Former Brilliant Member

$\angle ABC=180-80-20=80^\circ$ . Therefore, $\triangle ABC$ is an isosceles triangle with $BC=AC$ .

Draw equilateral $\triangle BCE$ and connect the points $A$ and $E$ as shown in my figure.

$\angle ABE=80-60=20^\circ$

Since $AB=DC,EB=BC$ and $\angle ABE = \angle BCD$ , $\triangle ABE \cong \triangle BCD$ $(S.A.S.)$ . Thus $\angle AEB=x$ .

Consider $\triangle ACE:$ $\angle ACE=60-20=40; \angle AEC = x+60;\angle CAE = x+60$

Finally,

$x+60+x+60+40=180$ $\implies$ $2x+160=180$ $\implies$ $2x=20$ $\implies$ $\boxed{x=10}$

Define BD = d. <BDC = 160 -x, so < BDA = 20 + x, and <ABD = 80 - x. By the Law of sines, d/sin(80) = AB/sin(20 + x). Also, d/sin(20) = DC / sin(x). But d/AB = d/ DC, so sin(80)/sin(20 + x) = sin(20)/ sin(x).Expanding, and simplifying, .342020143cos(x) = 1.939692621sin(x), or tan(x) = .176326981,and x = 10.0 Ed Gray