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

Draw BC = 8 cm

Draw the perpendicular at B and cut BA = 6 cm on it. Join AC right ΔABC is obtained.

Draw BD perpendicular to AC.

Since ∠BDC = 90° and the circle has to pass through B, C and D. BC must be a diameter of this circle. So, take O as the midpoint of BC and with O as centre and OB as radius draw a circle that will pass through B, C and D.

To draw tangents from A to the circle with centre

Join OA, and draw its perpendicular bisectors to intersect OA at

With E as centre and EA as radius draw a circle that intersects the previous circle at B and F.

Join AF.

Proof:
∠ABO = ∠AFO = 90° (Angle in a semicircle)
∴ AB ⊥ OB and AF ⊥ OF
Hence AB and AF are the tangents from A to the circle with centre O.
Video Solution:
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is perpendicular to AC. The circle through B, C and D is drawn. Construct the tangents from A to this circle
NCERT Solutions Class 10 Maths  Chapter 11 Exercise 11.2 Question 6:
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is perpendicular to AC. The circle through B, C and D is drawn. Construct the tangents from A to this circle
AB and AF are the required tangents from vertex A of triangle ABC to the circle with centre O