Question

In the picture, PQ is the diameter of the circle and O is the center of the circle. A tangent line is drawn from a point A which is on circle this tangent line touches the increased portion of PQ at R. Then prove this which is shown in attached pic plzz solve it friends it's urgent.

Answer 1

Since RA is the tangent,

Angle QAR = Angle QPA = x(say) (Angle in the alternate segment)

Angle QAP = 90° (Angle in a semi circle)

Therefore angle AQP = 180° - 90° - x = 90° - x.

Therefore Angle AQR = 180° -(90° - x) = 90° + x.

Now angle QRA = 180° -(90° + x) + x

Angle QRA = 90° - 2x

2x=90°- angle QRA

x= 1/2 (90° - angle QRA)

Angle QAR = 1/2 (90° - angle QRA)

Proved.