a)Let us suppose that $AL$ bisects the $\angle HAD$ so we get that $AH$ is a symmedian.From Steiner's Theorem we get $\frac{AB^2}{AC^2}=\frac{BH}{HC}$.We note $\angle BAH=\gamma$ and $\angle HAC=\beta$.We rewrite Steiner's relation and we get $\frac{AB}{sin\gamma}=\frac{AC}{sin\beta}$ so we get that $\frac{sin\beta}{sin\gamma}=\frac{cos\gamma}{cos\beta}$.Afther cross multiplying we get that $\sin2\beta=sin2\gamma$ from here we conlcude that either $\beta=\gamma$ which is nonsense either $2\beta+2\gamma=180$ and so $\beta+\gamma=\angle BAC=90$. So we are done. b)By easy angle chasing and using the fact that $D$ will be the center of circumcercle and so $\angle ADC=2\angle BCA$ we will easy solve the problem. P.S.Sorry for bad explanation but I'am hurring now.

Proof of $\angle BAC = 90^{\circ} \implies \angle HAL = \angle DAL$ Let $\angle C=\angle BCA = \alpha$ $BD=DC=AD \implies \angle DAC = \alpha$ $\implies \angle DAL=45^{\circ}-\alpha \text{ } (\text{ Note that } \alpha<45^{\circ})$ $\text{Also } \angle ABC= 90^{\circ}-\alpha$ $\implies \angle BAH = \alpha$ $\implies \angle LAH= 45^{\circ}-\alpha=\angle DAL$

The problem is nice , maybe we can use a pure geometric way to prove it : extend $AD$ to $E$ such that $AD=DE$ , extend $AH$ to $F$ such that $AH=HF$ , so we get $ABEC$ is a parallelogram , and we get $AB=EC=FB$ ,$ BC // EF $ ,notice that $BF$ is not parallel to $CE$ , we know that $CEFB$ is a isosceles trapezium , hence $C,E,F,B$ are concyclic , because $ \angle HAL=\angle DAL $ , so $\angle DAC=\angle DEB=\angle HAB=\angle HFB$ , hence $A,C,F,B$ are concyclic , so the five points $A.C.E.F.B$ are concyclic , parallelogram $ABEC$ is inscribed in a circle , so it should be a rectangle , hence $\angle BAC=\frac{\pi}{2}$ , If $\angle BAC=\frac{\pi}{2}$ , by easy calculation , we can get the result .

Here is another solution to this nice problem: We can easily observe that $AM$ is the isogonal conjugate of $AH$ but it is whell known that the isogonal conjugate of an altitude is the line $AO$(it is being proved by extremely easy angle chasing) so we get that $A,O$ and $M$ are collinear,if the $\triangle ABC$ is not isosceles then there is only one posibility that this is true and that is when $O$ and $M$ coincide so $\angle BAC=90$ So we are done.

1)Showing $\angle HAL=\angle DAL \implies\angle BAC=90^{\circ}$: Let $d$ and $l$ be tangents to circle $(ABC)$ from $B$ and $C$, $d\cap l=\{T\}$. And $AH$ is clearly symedian. As it known $AH$ passes through $T$.Then $AT\perp BC$. Clearly $DT\perp BC \implies AT\| DT$ . $\angle B \neq \angle C \implies AT\neq DT$. This means $T$ is point at infinity. From that $d\perp BC \implies\angle TBC=\angle A=90^{\circ}$ (If you don't want to use "point at infinity" you can say "$AT\| DT\implies$ $T$ doesn't exist. Then $d$ and $l$ should be parallel to $AH$. That means $d\perp BC\implies \angle A=90^{\circ}$) 2)Showing $\angle BAC = 90^{\circ} \implies \angle HAL=\angle DAL$: From $|AD|=|DC|$ we can say $\angle DAC=\angle ACD=\angle BAH$. Since $AL$ is angle bisector $\angle HAL=\angle DAL$

n a triangle $\triangle ABC$ with$\angle B>\angle C$, the altitude, the angle bisector, and the median from $A$ intersect $BC$ at $H, L$ and $D$, respectively. Show that $\angle HAL=\angle DAL$ if and only if $\angle BAC=90^{\circ}$.