Pg 222 and question no. 6 of ncert

Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle. Give the justification of the construction.

while doing this question i facing problem in drawing the line BD

the ncert solutions on this site are also not helpfull(so please dont copy and paste the answer that is on this site)

D is the point where the line segment AC intersects the circle. Join BD. 

As BD ⊥ AC, ∴∠BDC = 90°. Thus, if a circle is drawn through B, D, and C, BC will be its diameter. The centre E of this circle is the mid-point of BC.

The required tangents can be constructed on the given circle as follows.

1) Join AE and bisect it. Let F be the mid-point of AE.

2) Taking F as centre and FE as radius, draw a circle which will intersect the circle at point B and G. Join AG.

AB and AG are the required tangents.

Justification of construction: The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.

∠AGE is angle in the semi-circle. 

∴ ∠AGE = 90° (angle in a semi-circle is a right angle)

⇒ EG ⊥ AG

Since EG is the radius of the circle, AG has to be a tangent of the circle.

But we have proved ∠B = 90°

⇒ AB ⊥ BE

Since BE is the radius of the circle, AB has to be a tangent of the circle.