Question 1107835
<br>
I used coordinate geometry to solve the problem; there are certain to be other paths, possibly much easier....<br>
Let points C and D be C(0,0) and D(1,0).<br>
Consider the arc in the first quadrant of the circle with center C and radius CD.  To make AC=DC, point A must be somewhere on that arc.<br>
The requirement is that the height h of the trapezoid be equal to the length of base AB; so we need to find the coordinates of the point A on the arc such that the y coordinate (the height of the trapezoid) is equal to the length of base AB.<br>
Let h be the height of the trapezoid.<br>
By symmetry, the midpoint of base AB will have coordinates (0.5,h).<br>
Again by symmetry, the coordinates of point A will be (0.5+0.5h,h).<br>
Then since point A is on an arc of a circle with radius 1,<br>
{{{(0.5+0.5h)^2 + h^2 = 1}}}
{{{(1+h)^2+4h^2 = 4}}}
{{{1+2h+h^2+4h^2 = 4}}}
{{{5h^2+2h-3 = 0}}}
{{{(5h-3)(h+1) = 0}}}
{{{h = 3/5}}}  or  {{{h = -1}}}<br>
Clearly we need to choose the positive solution.<br>
So h, the length of base AB, is 3/5; then since the length of base CD is 1, the ratio of the lengths of the bases AB:CD is 3:5.<br><br>
------------------------------------------------------<br>
Having solved the problem by that method, I see that the right triangle formed by AC, DC, and the altitude from A to CD is a 3:4:5 right triangle; that suggests to me another possibly easier solution.<br>
Use the same figure as before; and let BE and AF be altitudes of the trapezoid.<br>
Let CE = DF = x and CD = y.  Then in right triangle ACF,
{{{CF = y-x}}}
{{{AF = EF = y-2x}}}
{{{AC = y}}}<br>
Then<br>
{{{(y-x)^2+(y-2x)^2 = y^2}}}
{{{y^2-2xy+x^2+y^2-4xy+4x^2 = y^2}}}
{{{y^2-6xy+5x^2 = 0}}}
{{{(y-x)(y-5x) = 0}}}<br>
Clearly y=x does not make sense in the problem; so y = 5x.<br>
But that makes the lengths of the sides of right triangle ACF 3x, 4x, and 5x.<br>
And since AB is the height of the trapezoid, the ratio AB:CD of the lengths of bases of the trapezoid is 3x:5x, or 3:5.<br>
Fun problem....!