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, and D is drawn. Construct the tangents from A to this circle. Give the justification of the construction.

Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as ∠BDC is of measure 90°. 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.

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