In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC
Solution:
Given, ABC is a right triangle with B at right angle.
A circle is drawn with AB as diameter intersecting the hypotenuse AC at P.
We have to prove that the tangent to the circle at P bisects BC.
We know that angle in a semicircle is always equal to 90°
So, ∠APB = 90°
A linear pair of angles is formed when two lines intersect.
By linear pair of angles,
∠BPC = 90°
So, ∠3 + ∠4 = 90° -------------------- (1)
Given, ∠B = 90°
In triangle ABC,
We know that the sum of all three interior angles of a triangle is always equal to 180°
∠BAC + ∠ABC + ∠ACB = 180°
∠1 + 90° + ∠5 = 180°
∠1 + ∠5 = 180° - 90°
∠1 + ∠5 = 90° -------------------------- (2)
We know that the angle between the tangent and the chord of a circle is equal to the angle made by the chord in the alternate segment.
So, ∠1 = ∠3 ------------------ (3)
Substitute (3) in (2),
∠3 + ∠5 = 90° --------------- (4)
Comparing (1) and (4),
∠3 + ∠4 = ∠3 + ∠5
∠3 + ∠4 - ∠3 = ∠5
∠4 = ∠5
From the figure,
∠4 = ∠CPQ
∠5 = ∠PCQ
So, ∠CPQ = ∠PCQ
We know that the sides opposite to equal angles are equal.
QC = PQ ------------------------ (5)
We know that the tangents drawn through an external point to a circle are equal.
Now, PQ = BQ
From (5), BQ = QC
This implies that PQ bisects BC
Therefore, it is proved that the tangent to the circle at P bisects BC.
✦ Try This: ABC and DBC are two right triangles with common hypotenuse BC and with their sides, AC and DB intersecting at P. Prove that AP. PC = BP . PD.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 6
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC
Summary:
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. It is proven that the tangent to the circle at P bisects BC
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