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

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 semi-circle)

∴ 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.