Calculate the smallest possible value to measure the ∠BAC angle Let $ABC$ be a right triangle acutangle and let $\overline{ AD}$, with D in $\overline{BC}$, be a height relative to point $A$. Let $Γ_1$
and $Γ_2$ as circumferences circumscribed to triangles $ABD$ and $ACD$, respectively. The circumference
$Γ_1$ crosses the $AC$ side at points $ A$ and $P$, while $Γ_2$ crosses the AB side at points $B$ and $Q$. Let
$X$ or line intersection point
BP with $Γ_2$ so that $P$ is between $B$ and $X$. Likewise, be $Y$ the intersection point of the line
$QC$ with $Γ_1$ so that $ Q $ is between $C$ and $Y$. Knowing that $A, X$ and $Y$
 are collinear, calculate the smallest possible value to measure the $\angle{BAC}$ angle.
I think triangular inequality would help.
I'm not sure about this condition
Problem wrote:
''Let ABC be a right triangle...''
If $A=90^{\circ}$ then $P=D$ or else $D=B$ or $C$. 


In this optimized drawing we have:
BA=BC=CY
AX=AY
$\angle BPA=90^o$
$\angle BQY=90^o$
BC=a, AC=b, AB=c
In triangle ACX, AC is diameter of circle so $\angle CXA=90^o$ and we can write:
$2AX^2=AC^2=b^2$
$AX \times YX=AX \times 2 AX=2AX^2$
⇒ $AX \times YX= b^2$
$QY \times QC = AX \times YX$
$QC=b Cos(\angle QCA)=b Sin(\angle CAQ)$
$CY=AB=BC=a$
⇒ $CY \times QY =a(a- CQ)=a(a-b Sin (\angle CAQ))=b^2$
Let $\angle CAQ=\angle CAB=\alpha$
⇒ $Sin (\alpha)= \frac{a^2-b^2}{ab}=\frac{a}{b}-\frac{b}{a}$ .         .         .        .      .       .     .     .       (1)
In triangle ABC we can write:
$Sin (\angle CBA)=Sin 2(PBA)=2Sin(90-CAB)Cos(90-CAB))= 2 Sin(CAB)Cos(CAB)=Sin 2(CAB)$
$\frac{b}{Sin (\angle CBA)}=\frac{a}{Sin (\alpha)}$
⇒ $\frac{b}{Sin 2(CAB)}=\frac{a}{Sin (CAB)}$
⇒ $\frac{b}{2 Cos (\alpha)}=a$
⇒  $\frac{b}{a}=2 Cos (\alpha)$
Putting this in relation (1) we get:
$Sin (\alpha)=\frac{1}{2 Cos(\alpha)}-2 Cos (\alpha)= \frac{1- 4 Cos^2(\alpha)}{2 Cos(\alpha)}$
⇒ $2Cos^2 (\alpha)=1-4 Cos^2(\alpha)$
⇒ $Cos^2(\alpha)=\frac{1}{6}$
⇒ $Cos (\alpha)≈ 0.4$ ⇒ $\alpha ≈ 66^o $
