In a parallelogram $ABCD$, taking $C$ as center and $CD$ as radius, draw a circle that intersects $BC$ at $N$ and contains the center of the parallelogram. If $AB=3$ is $AD=5$, calculate DN. (S:$2\sqrt{\frac{21}{5}}$)
I try:
$R=CE=CN=DC=3$
$CE = AE = 3$
$\angle DAC=\angle BCA \implies 3^2=6^2+5^2-2.6.5.cos \angle A \therefore cos \angle DAC =\frac{13}{15}$
$5^2=6^2+3^2-2.6.3..cos \angle ACD \implies cos \angle ACD = \frac{5}{9}$
$DE^2 = 3^2+5^2-2.3.5.cos \angle DAC \implies DE = 2\sqrt2 =BE$
On
Considering the figure we have:
$p_{ACD}=\frac{3+5+6}2=7$
$S_{ACD}=\sqrt{7(7-3)(7-5)(7-6)}=2\sqrt{14}$
$h=DE=\frac{2\sqrt{14}}{5/2}=\frac{4\sqrt{14}}5$
In triangle CDE we have:
$CE=\sqrt{CD^2-DE^2}=\sqrt{3^2-\frac{16\times 14}{25}}=\frac 15$
$NE=3-\frac 15=\frac {14}5$
$ND=\sqrt{h^2+NE^2}=\sqrt{(\frac{14}5)^2+(\frac{4\sqrt{14}}5)^2}=\frac{\sqrt{420}}5=2\sqrt{\frac{21}5}\approx 4.1$
where your answer is about 2 which is not correct.
You can use the law of cosines on $\triangle ACD$ to find $\cos\angle ADC$. Since $\angle DCN+\angle ADC=\pi$, we have $\cos\angle DCN=-\cos\angle ADC$. Finally, use the law of cosines on $\triangle DCN$ to find $DN$.
In addition, I doubt that the provided answer $DN=\sqrt{\frac{21}5}$ is correct. From your figure it seems that $DN>3$, but $\sqrt{\frac{21}5}\approx2.05$.