In the given figure, the diameter AB of the circle with centre O is extended to a point P and PQ is a tangent to the circle at the point T. If BPT = x and ATP = y, then prove that x + 2y = 90o.
It is known that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
OTP = 90o
In a triangle, the measure of an exterior angle is equal to the sum of the measures of its interior opposite angles.
Therefore, in PAT,
OAT = APT + ATP
OAT = x + y
In OAT, OA = OT (Radii of the same circle)
OAT = OTA
x + y = OTP - ATP
x + y = 90o - y
x + 2y = 90o
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