# 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

**☛ Related Questions:**

- In Fig. 9.18, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn par . . . .
- AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C . . . .
- Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining . . . .

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