# Thread: Geometry Question

1. ## Geometry Question

Let ABCD be a parallelogram with M as the midpoint of BC and N as the midpoint of CD. AngleNAM=60. AM=2. AN=1. Find AB.

I got that AMN is a 30-60-90, but where do I go from there?

2. ## Re: Geometry Question

In a 30-60-90 triangle, the side opposite the 30 degree angle is half the length of the hypotenuse and the side opposite the 60 degree side has length the square root of three times the length of the side opposite the 30 degree angle.

3. ## Re: Geometry Question

Yes, but that does not find a value for AB. I get that NM is sqrt3.

4. ## Re: Geometry Question

Let $\theta$ be angle ABC (which is equal to angle CDA). Then, suppose line segment BM has length $a$ and line segment CN has length $b$. Then AB has length $2b$ and AD has length $2a$. Next, apply the law of cosines to triangles DAN and BAM. You have:

$1^2 = (2a)^2 + b^2 - 2(2a)(b)\cos \theta$

$2^2 = a^2 + (2b)^2 -2(a)(2b)\cos \theta$

Subtracting the first equation from the second, you get $b^2 = 1+a^2$. Next, consider angle BCD. Because you have a parallelogram, it is $180^\circ - \theta$. Apply the law of cosines to triangle MCN.

$3 = a^2+b^2 - 2ab\cos(180^\circ - \theta) = a^2+b^2+2ab\cos \theta$

Multiplying both sides by 2, you have:
$6 = 2a^2+2b^2+4ab\cos\theta$

Adding this to the equation above: $1 = 4a^2+b^2-4ab\cos\theta$, you have:

$7 = 6a^2+3b^2$

Above, we found $b^2 = 1+a^2$, so plugging that in, you have:

$7 = 3+9a^2$ which implies $a^2 = \dfrac{4}{9}$, so $b^2 = 1+a^2 = \dfrac{13}{9}$. Then $b = \dfrac{\sqrt{13}}{3}$ and AB has length $2b = \dfrac{2\sqrt{13}}{3}$.