There are a lot of ways to solve it and here is one of them. Let $BD=2,DE=4,EC=2\sqrt{3}$ and we obtain $AB^2=6(\sqrt{3}+1)^2=(BE+DE)\cdot (DE+EC)=BE\cdot DC$,and hence $\frac{AB}{BE}=\frac{DC}{CA}$,together with $\angle ABE=\angle DCA=45^\circ$.It follows that $\triangle ABE\sim \triangle DCA$ and therefore,$\angle EAB=\angle BAD+\angle DAE=\angle CDA=\angle BAD+\angle ABC\implies \angle DAE=45^\circ$,as desired. $Q.E.D.$

For a generalization see RMO 2003 problem 1. Bye

Just apply cosine law in $DAC$ & in $BAE$ and then after getting all three sides of $EAD$, apply cosine law in $EAD$ to get $DAE$$=$$45$°.

Rotating the configuraiton about $A $by $90$◦, the point $B$ goes to the point $C$. Let $P$ denote the image of the point $D$ under this rotation. Then $CP = BD$ and ∠$ACP$ =∠$ABC$ = $45$◦, so $ECP$ is a right-angled triangle with $CE : CP$ = $\sqrt3 : 1$. Hence $P E = ED.$ It follows that $ADEP$ is a kite with $AP = AD$ and $P E = ED$. Therefore $AE$ is the angular bisector of ∠$P AD$. This implies that ∠$DAE$ = ∠$PAD/2$ = $45$◦

aayush-srivastava wrote: Rotating the configuraiton about $A $by $90$◦, the point $B$ goes to the point $C$. Let $P$ denote the image of the point $D$ under this rotation. Then $CP = BD$ and ∠$ACP$ =∠$ABC$ = $45$◦, so $ECP$ is a right-angled triangle with $CE : CP$ = $\sqrt3 : 1$. Hence $P E = ED.$ It follows that $ADEP$ is a kite with $AP = AD$ and $P E = ED$. Therefore $AE$ is the angular bisector of ∠$P AD$. This implies that ∠$DAE$ = ∠$PAD/2$ = $45$◦ Can you explain a bit more plz..

If we use sine law in this problem , will we get full marks ?

Yes of course we will get full marks