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.