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.
 A: Let $AD_1=x$ and $D_1B=a$
We are given that $AC^2=3BD^2$. Also note that $BC_1=AD_1=x$ because triangles $ADD_1$ and $BCC_1$ are congruent. By the Pythagorean theorem,
$$(2x+a)^2+(\sqrt{3}x)^2=3((\sqrt{3}x)^2+a^2)$$
$$2x^2+2a^2-4ax=0$$
$$(x-a)^2=0$$
so $x=a$. Now you have sufficient information to prove that ABD is an equilateral triangle.
A: If trig permissible, let $ a \neq b$ at start. Using CosineRule, minor diagonal
$$ a^2+b^2- 2ab\cdot  \frac12 = 1$$
and major diagonal
$$ a^2+b^2+2ab\cdot  \frac12 = 3 $$
Add and subtract,
$$ a^2+ b^2= 2,\quad 2ab =2; $$
Solve
$$ a=b=1 $$
So the parallologram has become its special case rhombus.
A: 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 $AB = AD_1 + D_1B =1$.
A: Let $\delta = \angle ADB$ and $\beta = \angle ABD$. It is enough to show that one of them is equal to $60°$.
For convenience, let $a=AD, b=AB, e = AC$, and $f=BD$.So, the parallelogram identity and law of sines give
$$2(a^2+b^2)=e^2+f^2 \stackrel{\frac{e^2}{f^2}= 3}{\Rightarrow} 2\left(\frac{\sin^2\delta}{\sin^2 60°} + \frac{\sin^2\beta}{\sin^2 60°}\right) = 4$$
Hence, after rearranging and noting that $\beta = 180°-60°-\delta$ you get
$$\sin^2\delta + \sin^2(120°-\delta) = \frac 32$$
Using the double-angle formula and reducing angles results in
$$\cos 2\delta + \cos (60°-2\delta) =0$$
and, applying $\cos u + \cos v = 2\cos \frac{u+v}{2}\cos\frac{u-v}{2}$, we obtain
$$\cos (2\delta - 30°) = 0 \Rightarrow \boxed{\delta = 60°}$$
