ex.11.2 Q.6

let ABC be a right triangle in which AB=6cm,BC=8cm and angle B=90^{0}. BD is the perpendicular from B on AC. The circle through B,C,D is drawn.Construct the tangents from A to this circle.

plz sm1 explain...urgent 2mrw exam....

Consider the following situation. If a circle is drawn through the points B, D, and C, BC will be its diameter as ∠BDC is of measure 90° and BC is the hypotenuse. The centre E of this circle will be the mid-point of BC.

The required tangents can be constructed on the given circle as follows.

**Step 1**

Join AE and bisect it. Let F be the mid-point of AE.

**Step 2**

Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. Join AG.

AB and AG are the required tangents.

**Justification**

The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.

∠AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠AGE = 90°

⇒ EG ⊥ AG

Since EG is the radius of the circle, AG has to be a tangent of the circle.

Already, ∠B = 90°

⇒ AB ⊥ BE

Since BE is the radius of the circle, AB has to be a tangent of the circle.

**
**