Finding a diagonal in a trapezoid given the other diagonal and three sides The figure below is a trapezoid, what is the length of the red line?
Thank you very much in advance!

 A: There are 2 circles, one of radius $49$ , with center $(0,0)$the other with radius $28$ with center $(49,0)$, solving the system:
$$x^2+y^2=49^2$$
$$(x-49)^2+y^2=28^2$$
Gives you the intersection point $(41,12\sqrt{5})$. Then you solve the system
$$y=12\sqrt{5}$$
$$(x-49)^2+y^2=28^2$$
and cuts the circle with radius $28$ at $x=57$, but we have also $y=12\sqrt{5}$ so the lenght is $\sqrt{57^2+(12\sqrt{5})^2}=63$
A: Well, I think I see a method of proceeding.  Let $x$ be one of the 2 equal angles of the isoceles triangle to the left.  It's possible to use the law of cosines to calculate $\cos x$.
It turns out the left triangle is congruent to the right triangle.  At least if you assume the drawing is accurate enough that the top and bottom are the 2 parallel lines of the trapezoid and not left and right.  The 2 equal angles of the left and right isoceles triangles are equal due to alternate interior angles (had to look up the name, been a while).  So the third angle of the triangle on the right is $180^o-2x$ and the lower right angle of the trapezoid is $x+180^o-2x=180^o-x.$
Now you can use the law of cosines along with $\cos(180^o-x)=-\cos x$ to solve for the length of the red line.
Not sure it's as elegant as dot dot's answer, but let's make sure it gets the same result.  We get that
$$\cos x=\frac{49^2+28^2-49^2}{2\times49\times28}=\frac27$$
$$c^2=49^2+28^2-2\times49\times28\times-\frac27=$$
$$49\times7^2+16\times7^2+16\times7^2=81\times7^2$$
$$c=9\times7=63$$
A: Since the two triangles in figure are similar, the length of the upper base of the trapezoid is $28\cdot 28/49 = 16$, so the heigth is $\sqrt{28^2-8^2}=12\sqrt{5}$ and the length of the red line is $\sqrt{(12\sqrt{5})^2+(49+8)^2}=63.$
