A few days ago, I came across a problem that has driving me insane. It seems so simple, but I just can't solve it, it's like a blind spot to me. Here's the problem:
A point $N$ is taken on the side $AB$ of a rectangle $ABCD$, such that $ND=CD$. Find the measurement of $\angle CND$ if $AB=2BC$.
Thanks in advance!
On
Another approach:
In figure you want to find angle AE'B. Drop perpendicular E'F from E' in figure on AB, in right angle triangle E'FA we have:
$E'F=\frac 12 AE' \Rightarrow \angle E'AF=30^o$
Reflect AB and BE' about AE'. Connect F to F'. Draw a circle center on H passing E'F and a line parallel with AB, they intersect at I . I is midpoint of arc E'F, so F'I is bisector of angle FF'E'. we have
$\angle FF'E'=30^o \Rightarrow \angle FE'B=\angle FF'I=15^o$
$AE'F=60^o \Rightarrow AE'B=60 +15= 75^o$
Notice that $ND:DA=CD:DA=2:1$. This implies that $\angle AND=\arcsin\frac{1}{2}=30^\circ$. Let $\angle NCD=\angle CND=\alpha$. Then, $\angle BCN=90^\circ-\alpha$, implying that $\angle BNC=\alpha$. Hence, we have $2\alpha+30^\circ=180^\circ$. Solving this equation gives us $\boxed{\alpha=75^\circ}$.