Let $ABC$ be a right triangle in which $AB = 6 \, cm$,$BC = 8 \, cm$ and $\angle B = 90^{\circ}$. $BD$ is the perpendicular from $B$ on $AC$. The circle through $B, C, D$ is drawn. Construct the tangents from $A$ to this circle and provide the justification for the construction.

(N/A) Construction steps:
$1$. Construct $\triangle ABC$ with $AB = 6 \, cm$,$BC = 8 \, cm$ and $\angle B = 90^{\circ}$.
$2$. Draw $BD \perp AC$. Since $\angle BDC = 90^{\circ}$,the circle passing through $B, C, D$ will have $BC$ as its diameter because $\angle BDC = 90^{\circ}$ subtends a right angle at $D$.
$3$. Let $O$ be the midpoint of $BC$. Draw a circle with center $O$ and radius $OB = OC = 4 \, cm$. This circle passes through $B, C$ and $D$ (since $\angle BDC = 90^{\circ}$).
$4$. To construct tangents from $A$ to this circle,join $AO$. Bisect $AO$ at $M$. Taking $M$ as center and $MA$ as radius,draw a circle. Let this circle intersect the circle with center $O$ at $B$ and $E$. Join $AE$. Thus,$AB$ and $AE$ are the required tangents.
Justification:
Join $OE$. Since $AB$ is a tangent to the circle at $B$,$\angle ABO = 90^{\circ}$. Since $AE$ is a tangent to the circle at $E$,$\angle AEO = 90^{\circ}$. In the circle with center $O$,$OB$ and $OE$ are radii. Since $AB$ and $AE$ are tangents from an external point $A$,$AB = AE$.