geometry help

In convex quadrilateral abcd,ab=bc=13 , cd=da=24, and angle D= 60 degrees. 

Points X andY  are the midpoints of BC and  DA respectively. Compute XY^2 (the square of the length of  XY).

\(\text{Let $\angle DAC =\angle ACD = 60^\circ $} \\ \text{Let $\angle ACY = \dfrac{\angle ACD}{2} = 30^\circ $} \\ \text{Let $ AD=AC=CD =24 $} \\ \text{Let $ CY = u $} \)

\(\mathbf{u=\ ?}\)

\(\begin{array}{|rcll|} \hline u^2+12^2 &=& 24^2 \\ u^2 &=& 24^2-12^2 \\ \mathbf{u^2} &\mathbf{=}& \mathbf{432} \\ \hline \end{array}\)

cos-Rule \(\mathbf{\cos(B)=\ ?}\):

\(\begin{array}{|rcll|} \hline 24^2 &=& 13^2+13^2-2\cdot 13 \cdot 13 \cdot \cos(B) \\ \ldots \\ \mathbf{\cos(B)} &\mathbf{=}& \mathbf{1-\dfrac{24^2}{2\cdot 13^2}} \\ \hline \end{array}\)

\(\mathbf{\cos\left(\dfrac{B}{2}\right)=\ ?}: \)

\(\begin{array}{|rcll|} \hline \cos(B) &=& 2\cos^2\left(\dfrac{B}{2}\right) - 1 \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 1+ \cos(B) \quad | \quad \mathbf{\cos(B) = 1-\dfrac{24^2}{2\cdot 13^2} } \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 1+ 1-\dfrac{24^2}{2\cdot 13^2} \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 2-\dfrac{24^2}{2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& 1-\dfrac{24^2}{2^2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& \dfrac{2^2\cdot 13^2-24^2}{2^2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& \dfrac{10^2}{2^2\cdot 13^2} \\ \cos\left(\dfrac{B}{2}\right) &=& \dfrac{10}{2\cdot 13} \\ \mathbf{\cos\left(\dfrac{B}{2}\right)} &\mathbf{=}& \mathbf{\dfrac{5}{13}} \\ \hline \end{array}\)

\(\mathbf{\sin(\alpha)=\ ?}\):

\(\begin{array}{|rcll|} \hline 180^\circ &=& B + 2\alpha \\ \ldots \\ \alpha &=& 90^\circ -\dfrac{B}{2} \\ \sin(\alpha) &=& \sin\left( 90^\circ -\dfrac{B}{2} \right) \\ \sin(\alpha) &=& \cos\left(\dfrac{B}{2} \right) \\ \mathbf{\sin(\alpha)} &\mathbf{=}& \mathbf{\dfrac{5}{13}} \\\\ \cos(\alpha) &=& \sqrt{1-\sin^2(\alpha)} \\ &=& \sqrt{1-\dfrac{5^2}{13^2}} \\ &=& \sqrt{ \dfrac{13^2-5^2}{13^2}} \\ &=& \sqrt{ \dfrac{12^2}{13^2}} \\ \mathbf{\cos(\alpha)} &\mathbf{=}& \mathbf{\dfrac{12}{13}} \\ \hline \end{array}\)

cos-Rule \(\mathbf{\overline{XY}^2=\ ?}\):

\(\begin{array}{|rcll|} \hline \overline{XY}^2 &=& 6.5^2+u^2-2\cdot 6.5 \cdot u \cdot \cos(\alpha+30^\circ) \\ &=& 6.5^2+432-13 \cdot u \cdot \cos(\alpha+30^\circ) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \cos(\alpha+30^\circ) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \cos(\alpha) \cos(30^\circ)-\sin(\alpha)\sin(30^\circ ) \Big) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \dfrac{12}{13}\cdot \dfrac{\sqrt{3}} {2} -\dfrac{5}{13}\cdot \dfrac{1} {2} \Big) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \dfrac {12 \sqrt{3}-5}{2\cdot 13} \Big) \\ &=& 474.25- \sqrt{432} \cdot \Big( \dfrac {12 \sqrt{3}-5}{2 } \Big) \\ &=& 474.25- \sqrt{\dfrac{432}{4}} \cdot \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{108} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{4\cdot 27} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{2^2\cdot 3^2\cdot 3} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- 6\sqrt{ 3} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- 6\sqrt{ 3} \cdot 12 \sqrt{3} -5\cdot 6\sqrt{ 3} \\ &=& 474.25- 72\ \cdot 3 +30\sqrt{ 3} \\ &=& 474.25- 216 +30\sqrt{ 3} \\ &=& 258.25 + 30\sqrt{ 3} \\ &=& 258.25 + 51.9615242271 \\ \mathbf{\overline{XY}^2} &\mathbf{=} & \mathbf{310.211524227} \\ \hline \end{array}\)

This problem becomes easier if we lay it out like this :

Let D  = (0,0)

A  =  (-24 cos60, 24 sin 60)  =  (-12, 12√3)

C = (24 cos 60, 24 sin 60)  =  (12, 12√3)

Let B = (0, y)

To  find y....we can use the square of the distance formula

(12 - 0)^2  + (y - 12√3)^2  = 13^2

12^2 + (y - 12√3)^2  =  13^2

(y - 12√3)^2  = 13^2 - 12^2

(y - 12√3)^2  = 25       take the square root of both sides

y - 12√3  = 5

y =  = 5 + 12√3

So....B  = (0, 5 + 12√3)

X =  [  (12 + 0)/2,  (12√3 + 5 + 12√3)/2 ]  =   (6, 2.5 +12√3)

Y  =  (-6, 6√3)

So XY^2   =    (6 - - 6)^2  + ( 2.5 + 12√3  - 6√3)^2  =

(12)^2  + (2.5 + 6√3)^2  =

144 + 6.25 + 30√3 + 108  =

258.25 + 30√3  units  ≈ 310.212  units