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In a right triangle $ A B C $ in which $ \angle B=90^{\circ} $, a circle is drawn with $ A B $ as diameter intersecting the hypotenuse $ A C $ at $ P $. Prove that the tangent to the cirlle at $ P $ bisects BC.
Given:
In a right triangle \( A B C \) in which \( \angle B=90^{\circ} \), a circle is drawn with \( A B \) as diameter intersecting the hypotenuse \( A C \) at \( P \).
To do:
Prove that the tangent to the circle at \( P \) bisects BC.
Solution:
Let $O$ be the centre of the given circle.
Let the tangent at $P$ meets $BC$ at $Q$.
Join $BP$.
$\angle ABC = 90^o$ (Tangent at any point on the circle is perpendicular to the radius through the point of contact)
In $\triangle ABC$,
$\angle CAB + \angle BCA = 90^o$ (Angle sum property and $\angle ABC = 90^o$)
$\angle BPQ = \angle BAC$ (Angle between the tangent and the chord is equal to the angle made by the chord in alternate segment)
$\angle BPQ + \angle BCA = 90^o$ ……..(i)
$\angle APB = 90^o$ (Angle in semi circle)
$\angle BPQ + \angle CPQ = 90^o$ …….(ii) ($\angle APB + \angle BPC = 180^o$, Linear pair)
From equations (i) and (ii), we get,
$\angle BPQ + \angle BCP = \angle BPQ + \angle CPQ$
$\angle BCP = \angle CPQ$
$PQ = QC$ (sides opposite to equal angles are equal)
$QP = QB$ (Tangents drawn from an internal point to a circle are equal)
$QB = QC$
Hence proved.
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