Question 1060279
{{{drawing(300,300,-2,2,-0.5,3.5,
line(-1,0,-1.62,1.9),line(1,0,1.62,1.9),
line(0,3.08,-1.62,1.9),line(0,3.08,1.62,1.9),
triangle(1.62,1.9,1,2,1,0),triangle(1.62,1.9,0,3.08,1,0),
rectangle(-1,0,1,2),locate(-1.05,0,A),
locate(0.98,0,B),locate(1.65,2,C),
locate(-0.02,3.3,D),locate(-1.75,2,E),
locate(-0.96,2,Y),locate(0.88,2,X)
)}}} The sides of the pentagon and the sides of the square have all the same length,
so triangle BCX is isosceles, and DCB is isosceles.
{{{ABX=90^o}}} because ABXY is a square.
{{{ABC=108^o}}} because ABCDE is a pentagon,
with external angles measuring {{{360^o/5=72}}} ,
and internal angles measuring {{{180^o-72^o=108^o}}} .
So, {{{ABC=BCD=108^o}}} .
{{{CBX=ABC-ABX=108^o-90^o=highlight(18^o)}}}
Since BCX is isosceles, besides its {{{CBX=18^o}}} vertex angle,
it has two base angles measuring
{{{XCB=(180^o-CBX)/2=(180^o-18^o)/2=162^o/2=81^o}}} .
Since DCB is isosceles, besides its {{{BCD=108^o}}} vertex angle,
it has two base angles measuring
{{{CBD=(180^o-BCD)/2=(180^o-108^o)/2=72^o/2=36^o}}}
{{{DCX=BCD-XCB=108^o-81^o=highlight(27^o)}}} .
{{{XBD=CBD-CBX=36^o-18^o=highlight(18^o)}}}