Can you solve this RMO question?

Rotate $\Delta ABC$ clockwise by $90^\circ$ .

Then $CD'=AD=3$ , $CE=4$ , and $\angle ECD'=90^\circ$ . Therefore $ED'=5=ED$ .

Thus $DED'B$ is a kite with $\angle DBD'=90^\circ$ , so $\angle DBE=45^\circ$ .

Let M be the midpoint of BC. Without losing generality, let BC=12.
So BD=3, DM=3, ME=2, EC=4 and AM=6.
$Applying~ Sin~ Law~ to~, \\ \Delta ~~ADB~~ \dfrac 3 {Sin(45-\alpha)}=\dfrac {AD}{Sin45} ....(1)\\ \Delta ~~ADM~~ \dfrac 3 {Sin(\alpha)}=\dfrac {AD}{Sin90}....(2)\\ \Delta ~~AEM~~ \dfrac2 {Sin(\beta)}=\dfrac {AE}{Sin90} ....(3)\\ \Delta ~~AEC~~ \dfrac 4 {Sin(45-\beta)}=\dfrac {AE}{Sin45} ....(4)\\ From~(1),(2)~~ \dfrac 3 {Sin(45-\alpha)}=\dfrac{ \frac 3 {Sin(\alpha)}}{Sin45} \\\implies~Sin\alpha=Cos\alpha-Sin\alpha\\ \therefore ~Tan\alpha=\color{#D61F06}{\dfrac1 2 }. \\ From~(3),(4)~~ \dfrac {4} {Sin(45-\beta)} = \dfrac { \frac 2 {Sin(\beta)}} {Sin45}\\ \implies~Sin\beta=Cos\beta-2Sin\beta\\ \therefore ~Tan\beta=\color{#D61F06}{\dfrac1 3} .\\ \therefore~\angle~DAE=\alpha+\beta=Tan^{-1}\dfrac{\frac1 2 +\frac1 3 }{1-\frac 1 2*\frac1 3}=~~~~~~~~~~~~~~\Large \color{#3D99F6}{45^o}.$

Good problem:easy yet beautiful

Cosine law can also be used though solution will be little lengthy

Much simple than I thought so