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.
Written on March 29, 2018, 7:45 p.m.