# Let ABC be a righ

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.

Justification:

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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