Question

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. A circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer 1

Steps of Construction:
Step I: $\Delta ABC$ is drawn.
Step II: Perpendicular to AC is drawn to point B which intersected it at D.
Step III: With O as a center and OC as a radius, a circle is drawn. The circle through B, C, D is drawn.
Step IV: OA is joined and a circle is drawn with diameter OA which intersected the previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.
Justification:
$\angle OEA={90}^{\circ }$ (Angle in the semi-circle)
$\therefore OE\perp AE$
Therefore, OE is the radius of the circle then AE has to be a tangent of the circle.
Similarly, AB is another tangent to the circle.