Points on a square

Note that we must clearly have $AE$ intersect line segment $CD$ , because otherwise $\angle DEA=\angle AEC$ is a clear contradiction. Let $AE$ and $CD$ intersect at $F$ . Denote the length of the side of the square as $s$ .

Then, let $\angle DEA=\angle AEC=x$ , so $\angle EAD=2x$ , $\angle EDA=180-3x$ and hence $\angle EDC=90-3x$ . Now, $\angle EFD=90+2x$ , $\angle EFC=90-2x$ , and thus $\angle ECD=90+x$ . Also, since $\angle EDC>0$ , we get that $0<x<30$ .

By the law of sines on $\triangle AED$ , $\frac{ED}{\sin (2x)}=\frac{s}{\sin x}$ , and by the law of sines on $\triangle CED$ , $\frac{ED}{\sin (90+x)}=\frac{s}{\sin (2x)}$ . Hence, by rearranging these two equations, we get $\sin ^2 (2x)=\sin x \sin (90+x)$ . Since $0<x<30$ , we have $\sin (90+x)=\cos x$ , so $\sin ^2 (2x)=\sin x \cos x$ , so by the sine double-angle formula, $\sin (2x)=\frac 12$ . The function $f(x)=\sin (2x)$ is strictly increasing as $x$ increases from 0 to 30, so hence there is at most one solution to $\sin (2x)=\frac 12$ in the range from 0 to 30. Clearly $x=15$ is such a solution, so therefore $\angle ECD=90+15=\boxed{105}$ .

Let $\angle DEA = \angle AEC = \theta$ and $\angle DAE = 2\theta$ . Then $\angle DCE = \theta+90^\circ$ . Let the side of the square be $a$ . Applying the Sine Rule to $\Delta ADE$ $\frac{DE}{\sin2\theta} = \frac{DA}{\sin\theta}$ and hence $DE = 2a\cos\theta$ . Applying the Sine Rule to $\Delta CDE$ , $\frac{DE}{\sin(\theta+90^\circ)} = \frac{DC}{\sin2\theta}$ and hence $DE = \frac{a}{2\sin\theta}$ . Thus $2\sin2\theta = 4\sin\theta\cos\theta = 1$ , so $\sin2\theta = 1/2$ , so $\theta=15^\circ$ . Hence $\angle DCE = 105^\circ$ .

Let $DEA=x$ , and assume $AD=1$ . By the sine rule on EDA we get $\frac{1}{\sin{x}}=\frac{ED}{\sin{2x}}$ Chasing angles it is easy to find $ECD=90+x$ and then the sine rule on ECD gives $\frac{1}{\sin{2x}}=\frac{ED}{\sin{(90+x)}}$ . Equating the two expressions for ED, and using $\sin{(90+x)}=\cos{x}$ gives $\sin{2x}=\frac{1}{2}$ and hence $2x=30$ and $ECD=105$

Let, $\angle DEA=\angle AEC=\frac{1}{2} \angle EAD=\alpha$ and $AB=BC=CD=DA=a$ and $DE=x$

Of $\Delta ADE$ ,
$\angle EAD+\angle DEA+\angle ADE=180^\circ$
⇒ $\angle ADE=180^\circ-2\alpha-\alpha$ [ We know that, $\angle DEA=\frac{1}{2} \angle EAD=\alpha$ ]
⇒ $\angle ADE=180^\circ-3\alpha$
⇒ $\angle CDE=90^\circ-3\alpha$

Of $\Delta DCE$ ,
$\angle DCE+\angle CED+\angle CDE=180^\circ$
⇒ $\angle DCE=180^\circ-2\alpha-90^\circ+3\alpha$ [ We know that, $\angle CED=\angle AEC+\angle DEA=\alpha+\alpha=2\alpha$ and $\angle CDE=90^\circ-3\alpha$ ]
⇒ $\angle DCE=90^\circ+\alpha$

Of $\Delta ADE$ ,
$\frac{AD}{sin \angle AED}=\frac{DE}{sin \angle DAE}$
⇒ $\frac{a}{sin \alpha}=\frac{x}{sin 2\alpha}$
⇒ $\frac{a}{x}=\frac{sin \alpha}{sin 2\alpha}$ .............( $1$ )

Of $\Delta DCE$ ,
$\frac{CD}{sin \angle CED}=\frac{DE}{sin \angle DCE}$
⇒ $\frac{a}{sin 2\alpha}=\frac{x}{sin (90^\circ+\alpha)}$
⇒ $\frac{a}{x}=\frac{sin 2\alpha}{sin (90^\circ+\alpha)}$ ...............( $2$ )

By equating ( $1$ ) and ( $2$ ), we shall get,
$\frac{sin \alpha}{sin 2\alpha}=\frac{sin 2\alpha}{sin (90^\circ+\alpha)}$
⇒ $\frac{sin \alpha}{2.sin \alpha.cos \alpha}=\frac{sin 2\alpha}{cos \alpha}$
⇒ $sin 2\alpha=\frac{1}{2}$
⇒ $2\alpha=30^\circ$ [ We know that, $0^\circ$ $\leq$ $\alpha$ $\leq$ $90^\circ$ ]
⇒ $\alpha=15^\circ$

So,
$\angle DCE=90^\circ+\alpha$
⇒ $\angle DCE=90^\circ+15^\circ$
⇒ $\angle DCE=105^\circ$ [ ANSWER ]

Let $\angle DEA = x$ . Through angle chasing we see that $\angle ECD = 90 +x$ . Now, we use law of sines on Triangles DEC and DAE. From the first triangle we have that $\frac{DC}{sin2x} = \frac{DE}{sin(90+x)}$ . The second triangle gives us $\frac{DA}{sinx} = \frac{DE}{sin2x}$ . We know that DA = Dc since ABCD is a square so equating the two relations to each other gives us that $\frac{sinx}{sin2x} = \frac{sin2x}{sin(90+x)}$ . This becomes $\frac{1}{sin2x} = \frac{2cosx}{sin(90+x)}$ , which further simplifies to sin2x = $\frac{1}{2}$ . This means that sinx = 15. Thus $\angle ECD = 105$ .

Just apply Sine rule twice and we are done..isn't this problem over-rated?