In a right angle ΔABC is which B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at PQ bisects BC.



Given : In a right angle ΔABC is which B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Also PQ is a tangent at P


To Prove : PQ bisects BC i.e. BQ = QC


Proof :


APB = 90° [Angle in a semicircle is a right-angle]


BPC = 90° [Linear Pair ]


3 + 4 = 90 [1]


Now, ABC = 90°


So in ABC


ABC + BAC + ACB = 180°


90 + 1 + 5 = 180


1 + 5 = 90 [2]


Now ,


1 = 3[angle between tangent and the chord equals angle made by the chord in alternate segment]


Using this in [2] we have


3 + 5 = 90 [3]


From [1] and [3] we have


3 + 4 = 3 + 5


4 = 5


QC = PQ [Sides opposite to equal angles are equal]


But Also PQ = BQ [Tangents drawn from an external point to a circle are equal]


So, BQ = QC


i.e. PQ bisects BC .


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