QR is a tangent at Q. PR //AQ, where AQ is a chord through A, the end point of the diameter
AB and P is a centre. Prove that BR is tangent at B.

Given, P is the center of the circle and PA and PQ are radius. 
so, In traingle PAQ,
 PA=PQ=r
Therefore, ∠1=∠2 (equal sides have equal angles opposite to them)

Now as PR║AQ,
∠1=∠3 (corresponding angles)
and ∠2=∠4
But, ∠1=∠2 (as proved earlier)
therefore ∠3=∠4

Now inΔPQR and ΔPBR
PQ=PB=r
∠3=∠4 (as proved earlier)
PR=PR (common)

therefore, ΔPQR≡ΔPBR (by SAS rule)
so ∠PQR=∠PBR (corresponding angles of congurent traingle)
here as QR is a tangent, 
therefore ∠PQR=90°

Hence, ∠PBR=90° (as ∠PBR=∠PQR)
Therefore, BR is a tangent at B.