Q. 7

# In the figure, O

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

OAP = OBP = 90°

Now, OAP + APB + OBP + AOB = 360° [Angle sum property of quadrilaterals]

90° + APB + 90° + AOB = 360°

AOB = 360° - 180° - APB = 180° - APB.... (1)

Now, in Δ OAB, OA is equal to OB as both are radii.

OAB = OBA [In a triangle, angles opposite to equal sides are equal]

Now, on applying angle sum property of triangles in Δ AOB,

We obtain OAB + OBA + AOB = 180°

2 OAB + AOB = 180°

2 OAB + (180° - APB) = 180° [Using (1)]

2 OAB = APB

Thus, the given result is proved

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