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Consider a circle, with centre $O$. Draw a chord $AB$ and a tangent $SR$ to the circle at point $B$. Chord $AB$ subtends angles at points $P$ and $Q$ on the minor and major arcs, respectively.

Draw a diameter $BT$ and join $A$ to $T$.

Prove that $\hat{APB}=\hat{ABR}$ and $\hat{AQB}=\hat{ABS}$
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Let PBR= x = PAB , tan-chord theorem

let AQB=y= APB, subtended by same chord AB.

Then ABP= 180-90-x = 90-x, <s in triangle APB

hence APB(90)=ABR(90-x + x) = 90 ,proven

AQB = 90 , already gotten