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