In a right triangle ABC in which =90^ , a circle is drawn with AB…

Question

In a right triangle ABC in which $\angle\text{B}=90^{\circ},$ a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects
BC.

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Answer

Given: $\triangle\text{ABC}$ is right triangle in which $\angle\text{ABC}=90^{\circ}.$ A circle is drawn with side AB as diameter intersecting AC in P.
PQ is the tangent to the circle when intersects BC in Q.
Construction: join BP.
Proof: PQ and BQ are tangents drawn from an external point Q.
⇒ PQ = BQ....(i) [Length of tangents drawn from an external point to the circle are equal]
$\Rightarrow\angle\text{PBQ}=\angle\text{BPQ}$ [In a triangle, equal side have equal angles opposite to them]
As, it is the diameter of the circle.
$\therefore\angle\text{APB}=90^{\circ}$ [Angle in a semi-circle is 90°]
$\angle\text{APB}+\angle\text{BPC}=180^{\circ}$ [Linear pair]
$\Rightarrow\angle\text{BPC}=180^{\circ}-\angle\text{APB}=180^{\circ}-90^{\circ}=90^{\circ}$
In $\triangle\text{BPC},$
$\angle\text{BPC}+\angle\text{PBC}+\angle\text{PCB}=180^{\circ}$[Angle sum property]
$\Rightarrow\angle\text{PBC}+\angle\text{PCB}=180^{\circ}-90^{\circ}=90^{\circ}...\text{(ii)}$
Now,
$\angle\text{BPC}=90^{\circ}$
$\Rightarrow\angle\text{BPQ}+\angle\text{CPQ}=90^{\circ}...\text{(iii)}$
From (ii) and (iii), we get,
$\Rightarrow\angle\text{PBC}+\angle\text{PCB}=\angle\text{BPQ}+\angle\text{CPQ}$
$\Rightarrow\angle\text{PCQ}=\angle\text{CPQ}[\angle\text{BPQ}=\angle\text{PBQ}]$
In $\triangle\text{PQC},$
$\angle\text{PCQ}=\angle\text{PQC}$
$\therefore$ PQ = QC ...(iv)
From (i) and (iv), we get,
BQ = QC
Thus, tangent at P bisects th side BC.