Show a parallelogram with angle $60^\circ$ is a rhombus 
If $ABCD$ is a parallelogram with $\angle BAD=60 ^\circ$ and $\dfrac{AC^2}{BD^2}=\dfrac31$, show $ABCD$ is a rhombus.


We have the squares of $AC$ and $BD$ so MAYBE it is a good idea to construct right triangles. Let $DD_1\perp AB$ and $CC_1 \perp AB$. Now we have the right triangles $BD_1D$ and $AC_1C$ with hypotenuses $BD$ and $AC$, respectively. By the Pythagorean theorem we can get $AC^2=AC_1^2+CC_1^2$ and $BD^2=BD_1^2+DD_1^2$. This does not seem to help. Can you give me some hints? Thank you in advane! :)
I am trying to solve it by the Pythagorean theorem.

You should be right on target.
Let $AD=1$ be our unit length.
We want to prove $AB = 1$.
But that would make $\triangle ADD_1$ an isosceles triangle with a $60$ degree vertex, or in other words and equilateral triangle, so $AB=BD=1$.
Let $BD = x$.
We have:
$AD_1 = \frac 12$.  $DD_1= \frac{\sqrt 3}2$. (Because $\triangle ADD_1$ is a $30-60-90$ triangle.)
$D_1B = \sqrt{x^2-\frac 34}$. (By Pythagorean theorem)
$BC_1 = AD_1 =\frac 12$. (Because $ABCD$ is a parallelogram.)
And $AC_1 =AD_1 + D_1B+BC_1 = \sqrt{3x^2 -\frac 34}$.(By Pythagorean theorem)
So we have $\frac 12 + \sqrt{x^2-\frac 34} + \frac 12=\sqrt{x^2-\frac34}+1 = \sqrt{3x^2 -\frac 34}$
Which has solution $x = 1$
And so $D_1B= \sqrt{x^2-\frac 34}=\sqrt{1^2 -\frac 34}=\frac 12$ and $AB = AD_1 + D_1B =1$.
