Let ABC be a right triangle in which $AB = 6\ cm, BC = 8\ cm$ and $\angle B = 90^o$. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Given:
A right triangle $ABC$ in which $AB = 6\ cm, BC = 8\ cm$ and $\angle B = 90^o$.
To do:
We have to construct tangents from $A$ to this circle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 8\ cm$.
(ii) From $B$ draw an angle of $90^o$.
(iii) Draw an arc $BA = 6\ cm$ cutting the angle at $A$.
(iv) Join AC. $ABC$ is the required triangle.
(v) Draw perpendicular bisector of $BC$ cutting $BC$ at $M$.
(vi) Taking $M$ as centre and $BM$ as radius, draw a circle.
(vii) Taking $A$ as centre and $AB$ as radius draw an arc cutting the circle at $E$. Join $AE$.
$AB$ and $AE$ are the required tangents.
Justification:
$\angle ABC = 90^o$ (Given)
$OB$ is a radius of the circle.
This implies,
$AB$ is a tangent to the circle.
Also, $AE$ is a tangent to the circle.
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