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# 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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