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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, 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 and 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 the 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 O.

Join OA, and draw its perpendicular bisector to intersect OA at point E.

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

Join AF.

Thus, AF and AB are the required tangents to the circle with centre O.

Proof:
∠ABO = ∠AFO = 90° (Angle in a semicircle)
∴ AB ⊥ OB and AF ⊥ OF (We know that the line joining the centre of a circle to the tangent is always perpendicular)
Hence AB and AF are the tangents from A to the circle with centre O.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 11
Video Solution:
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, D is drawn. Construct the tangents from A to this circle.
NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2 Question 6
Summary:
ABC is 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. AB and AF are the required tangents from vertex A of triangle ABC to the circle with centre O.
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