Q. 64.0( 338 Votes )
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is theperpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Steps of construction:
i. Draw a line segment AB = 6 cm.
ii. Draw a right angle ∠ABC at point B, such that BC = 8 cm.
iii. Now, draw a perpendicular bisector of BC which will intersect it at P .
iv. Now P is a mid point of BC.
Taking P as a centre and BP as radius draw a circle.
v. Join A to the centre of circle i.e. P.
Make perpendicular bisector of AP.
Let Q be the mid point of AP.
vi. Taking Q as centre and AQ as a radius draw a circle.
vii. Now Both circles intersect each other at B and R.
Hence AB and AR are the required tangents.
We need to prove AB and AR are tangents.
Construction: Join PR.
As ARP is an angle on the semicircle BPR.
And angles in semicircles are of 90°.
∴ ∠ARP = 90°
⇒ AR ⊥ PR
And PR is the radius of circle,
From the theorem which states that tangent is perpendicular to the radius.
So AR has to to be tangent.
Similarly AB is a tangent.
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