Question

8. In a right triangle ABC a circle with a side AB as diameter
is drawn to intersect the hypotenuse AC in P. Prove that
the tangent to the circle at P bisects the side BC.
​

Answer 1

Answer:

P(2,-1), Q(3,4), R(-2,3) and S(-3,-2) be four points in a plane, then show

that PQRS is a rhombus and not a square.

Answer 2

Answer:

Let O be the center of the given circle. Suppose, the tangent at P meets BC at Q. Then join BP. Required to prove: BQ = QC Proof : ∠ABC = 90° [tangent at any point of circle is perpendicular to radius through the point of contact] In ∆ABC, ∠1 + ∠5 = 90° [angle sum property, ∠ABC = 90°] And, ∠3 = ∠1 [angle between tangent and the chord equals angle made by the chord in alternate segment] So, ∠3 + ∠5 = 90° ……..(i) Also, ∠APB = 90° [angle in semi-circle] ∠3 + ∠4 = 90° …….(ii) [∠APB + ∠BPC = 180°, linear pair] From (i) and (ii), we get ∠3 + ∠5 = ∠3 + ∠4 ∠5 = ∠4 ⇒ PQ = QC [sides opposite to equal angles are equal] Also, QP = QB [tangents drawn from an internal point to a circle are equal] ⇒ QB = QC – Hence proved.

Step-by-step explanation:

hope it helps

please mark me as brainliest