The parpendicular drawn on radius at the end point of radius of a circle will be a tangent to the circle at the end point of radius.

Explanation

Given: Let AB be a diametr of the circle with centre at O. AC and BD are two tangents at the end-points A and B of AB.
To prove: `AC||BD.`
Proof : In the circel with centre at O , AC is a tangent at A and OA is a radius through point of contact.
`therefore` OA is perpendicular to AC. [by theorem-40]
`therefore angle OAC=1` right angle ........(1)
Again, in the circle with centre at O, BD is a tangent at the point B and OB is a radius through point of contact.
`therefore OB botBD` [bytheorem-40]
`therefore angle OBD=1` right angle .......(2)
Now AB has intersected tow line segments AC and BD and two adjacent angles on the same side of the transversal AB are `angle BACand angle ABD.`
Now, `angle ABD+angle ABD=angle OAC+ angle OBD`
`=1` right angle + 1 right angle [from (1) and (2)]
=2 right angles
`therefore` AC and BD are paralled to each other, i.e., `AC||BD.`