**
Ex 11.2, 6
**

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.

Given AB = 6 cm, BC = 8 cm and ∠ B = 90°

and BD is perpendicular from B on AC.

We draw circle through B, C, D

Now, ∠ BDC = 90°

Since ∠ BDC is angle formed by chord BC in circle & its 90°

∴ BC is the diameter of circle.

Thus, center of circle will be the bisector of line BC

**
Steps of construction
**

- Bisect line BC. Let E be mid-point of BC. Thus, E is center of circle. We need to construct tangents from point A to the circle
- Join line AE and bisect it. Let M be mid point of AE
- Taking M as centre and AM as radius, draw a circle.
- Let it intersect the given circle at points B and P.
- Join AB and AP.

Thus, AB and AP are the required tangents

**
Justification
**
**
:
**

We need to
__
prove that
__
__
AG and AB are the tangents to the circle
__
.

Join EP.

APE is an angle in the semi-circle

of the blue circle

And we know that angle in a

semi-circle is a right angle.

∴ ∠EPA = 90°

⇒ OQ ⊥ PQ

Since EP is the radius of the circle,

AP has to be a tangent of the circle.

Also, given ∠ B = 90°

Since EB is the radius of the circle,

AB is a tangent of the circle.