First Let $B,K,C,L $ be con cyclic.. Extend $DC$ to point $P$ such that $DC=DP.$ And then we get $PAC$ is isosceles and $PKCL$ is parallelogram .Also we get $KPAD$ is a cyclic quadrilateral. And then it is very easy Proving it other way is quite obvious..

This is an interesting problem. Let $B,K,C,L$ lie on a circle. Let $AD$ meet $(ABC)$ again at $X$ and note that $DA\cdot DX=DB\cdot DC=DK\cdot DL$ and the points $A,K,X,L$ are concyclic. As $X$ is the Miquel's Point of $BCKL$, $\triangle XKL \sim \triangle XBC$ and it follows that $X,D,B,K$ and $X,D,C,L$ are concyclic. Thus, $AX$ is a diameter in $(AKL)$ and a median in triangle $AKL$; and as $AKL$ is non-isosceles, it follows that $D$ is the circumcenter of $AKL$ and so $\angle BAC=90^{\circ}$. All steps are reversible and the converse promptly holds.

If we go by considering $BKCL$ concyclic , just observe that $KL$ is the $A$ symmedian of $ABC$ . Also as $AD$ is perpendicular to $BC$ , by the theorem :- "In triangle $ABC$ , if $AD$ is the altitude from $A$ to $BC$ , and $ O$ is circumcenter of $ABC$ , then, angle $BAD$ = angle $CAO$ "... Using this property , we get that the circumcenter of $ABC$ lies on the $A$ median of the triangle $ABC$ . But this happens iff $ABC$ is either an equilateral triangle , an isosceles triangle , or a right angled triangle. As $ABC$ is given to be a scalene triangle , this implies angle $BAC = 90$ . Proving the other way round , i.e taking angle $A = 90$ is trivial by angle chasing .

anantmudgal09 wrote: This is an interesting problem. Yeah, interesting indeed . Suppose, $BKCL$ is concyclic Let $\angle ALK$ = $k$. Then, $\angle KBC$ = $(180-k)$. for this, $\angle ABD$ = $180$ - $\angle KBC$ = $ k $., $\angle DAK$= $\angle DAB$ = $90 $ - $ \angle ABD $ = $ 90-k $......(@) Let, $ O $ is the circumcentre of $ \triangle AKL $. Here, $ \triangle OAK $ is isosceles. Here, $ \angle AOK $=2 ($ \angle AKL $)= $ 2k $ & $ \angle OAK $= $ 90 -k $......(b) From, @ & (b) ,$\angle DAK$ = $\angle OAK $ , SO, $ O $ lies on line $ AD $. If we construct a perpendicular bisector of $ KL $ through $ D $ , it will go through $O$. So, $O$ & $D$ is same point. so, $KL$ is a diameter of $\odot AKL $. AS, $\angle KAL$ is inscribed in a semi-circle, so, $\angle KAL$ = $\angle BAL$= $\angle BAC$ =$90^{\circ}$.

Sorry for reviving, but for the only if part I have got a trigonometric solution without any construction....So posting it. Suppose, $BKCL$ be cyclic. Let $\angle ACB= a$. $\implies$ $\angle DAC= (\pi\ /2 - a)$. Let $\angle KLC= b$ $\implies$ $\angle KBC= b$ $\implies$ $\angle ABC= (\pi\ - b)$ $\implies$ $\angle BAD= (b- \pi\ /2)$. Now, apllying Sine Rule in $\triangle ADK$ and $\triangle ADL$ , we get $ AD / KD = -sin a/cos b$ and $AD / DL = sin b/ cos a$ respectively. $\implies\ sin 2a + sin 2b=0 \implies\ 2sin(a+b)sin(a-b)=0$. Now, as $\triangle ABC$ is scalene, $\angle ABC \neq\ \angle ACB$. So, $ a+b\neq\ \pi$. So, $cos (a-b)=0 $. So, $|a-b|=\pi\ /2$. Now, as $BC$ is largest side, $\angle ABC$ and $\ angle ACB$ both must be acute, hence, $\angle KBC=b \ > \pi\ /2 > \angle ACB=a$ . So, $b-a=\pi\ /2$. so $a= b - \pi\ /2$, hence, $\angle DKA= \angle DAK$. So, $D$ is the circumcenter of $\triangle AKL$ .Hence $\angle BAC=\pi\ /2$.

If part: In $\Delta ABC$ if $\angle BAC=\frac{\pi}{2}$, then $AD^2=BD.DC$, and $AD$ being the median in right angled $\Delta AKL$ we have $AD^2=KD^2=KD.DL$ (since $AD=KD=DL$), and we get $BD.DC=KD.DL$ i.e. $\frac{BD}{KD}=\frac{DL}{DC}$, which means that the two triangles $\Delta BKD$ and $\Delta LCD$ are similar. This gives $\angle BKD=\angle LCD$. Only If part: If $BKCL$ is concyclic, then $\angle AKL=\angle ACB$ and $\angle LAB$ being the common angle, $\Delta AKL$ and $\Delta ABC$ are similar. Drawing the median $AM$ in $\Delta ABC$, so that $M$ is the midpoint of $BC$, we have that $AM$ and $AD$ play the same corresponding roles (of the median from the equal angle $A$) in the two similar triangles, and $AC$ and $AK$ play the same corresponding roles (of being the sides opposite to equal angles $\angle ALK=\angle ABC$ , since the two are similar.) So, the angle between $AM$ and $AC$ in $\Delta ABC$ must be equal to the angle between $AD$ and $AK$ in $\Delta AKL$. In particular, $\angle MAC=\angle DAB$, which means $AM$(median from $A$ in $\Delta ABC$) is the isogonal conjugate of $AD$(altitude from $A$ in $\Delta ABC$) in $\Delta ABC$. But we also know, that the isogonal conjugate of $AD$ is $AO$ where $O$ is the circumcenter of $\Delta ABC$. So, $AO$ and $AM$ is the same line, which means that essentially $M$ and $O$ are the same point. So $M$ is the circumcenter of $\Delta ABC$, which means that $\angle BAC= \frac{\pi}{2}$