Question 147598
{{{drawing( 300, 300, -1, 3, -1, 3,locate(-.2,0,A),locate(.5,1.3,B),locate(2.1,0,D),locate(1.4,1.3,C),line(0,0,2,0),line( 0, 0, .5, 1),line(0.5,1,1.5,1),line(1.5,1,2,0))}}}
Length of AB = x
Length of BC = x
Length of CD = x
Length of AD = 2x
Draw two perpendicular lines from BC to AD at points B and C.
{{{drawing( 300, 300, -1, 3, -1, 3,locate(-.2,0,A),locate(.5,1.3,B),locate(2.1,0,D),locate(1.4,1.3,C),green(line(.5,1,.5,0)),locate(0.5,-0.1,G),green(line(1.5,1,1.5,0)),locate(1.5,-0.1,H),line(0,0,2,0),line( 0, 0, .5, 1),line(0.5,1,1.5,1),line(1.5,1,2,0))}}}
These create new line BG and CH, and break up AD into AG, GH, and HD.
The triangles ABG and CDH are right triangles and from the diagram BC=GH=x.
{{{BG=CH}}}
{{{AG=HD}}}
{{{AD=AG+GH+HD}}}
{{{2x=AG+x+HD}}}
{{{2x=AG+x+AG}}}
{{{AG=x/2}}}
Now you have a right triangle with hypotenuse of x and an adjacent leg of x/2.
From trigonometry,
{{{cos(A)=Adj/Hyp}}}
{{{cos(A)=(x/2)/x}}}
{{{cos(A)=1/2}}}
{{{A=60}}}
Since ABG and CDH are identical, angle A and angle D are the same.
{{{D=60}}}