Q. 164.3( 23 Votes )

In the given figure, O is the center of the circle and TP is the tangent to the circle from an external point T. If PBT = 30°, prove that BA : AT = 2 : 1.


Answer :

In Given Figure, we have a circle with center O let the radius of circle be r.


Construction : Join OP



Now, In APB


ABP = 30°


APB = 90°


[Angle in a semicircle is a right angle]


By angle sum Property of triangle,


ABP + APB + PAB = 180


30° + 90° + PAB = 180


PAB = 60°


OP = OA = r [radii]


PAB = OPA = 60°


[Angles opposite to equal sides are equal]


By angle sum Property of triangle


OPA + OAP + AOP = 180°


60° + PAB + AOP = 180


60 + 60 + AOP = 180


AOP = 60°


As all angles of OPA equals to 60°, OPA is an equilateral triangle


So, we have, OP = OA = PA = r [1]


OPT = 90°


[Tangent at any point on the circle is perpendicular to the radius through point of contact]


OPA + APT = 90


60 + APT = 90


APT = 30°


Also,


PAB + PAT = 180° [Linear pair]


60° + PAT = 180°


PAT = 120°


In APT


APT + PAT + PTA = 180°


30° + 120° + PTA = 180°


PTA = 30°


So,


We have


APT = PTA = 30°


AT = PA


[Sides opposite to equal angles are equal]


AT = r [From 1] [2]


Now,


AB = OA + OB = r + r = 2r [3]


From [2] and [3]


AB : AT = 2r : r = 2 : 1


Hence Proved !


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