Q. 64.0( 305 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.

Answer :

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.
Join AR.

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.
Hence proved

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