Given, P is the center of the circle. OR is tangent to the circle at Q.

Again, AB is the diameter of the circle and AQ || PR.

From the figure,

In ΔAPQ,

AP = PQ     {Radius of the circle}

∠1 = ∠2   {since in a triangle, equal sides have equal angles opposite to them}

Again given,

AQ || PR

So, ∠1 = ∠3  {pair of corresponding angles}

     ∠2 = ∠4  {pair of alternate angles}

So,  ∠3 = ∠4  ( since  ∠1 = ∠2 )

Now, from Δ PQR and Δ PBR

PQ = PB  {radius of the circle}

 ∠3 = ∠4    {Proved}

PR = PR  {common}

By SSS congrience criterian,

Δ PQR ≅ Δ PBR

So, by CPCT

∠ PQR = ∠ PBR

Now, ∠ PQR = 90     {since radius is perpendicular to the tangent at the point of contact}

So, ∠ PBR = 90

Hence, BR is a tangent to the circle.