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# in a right triangle abc right angled at b a circle is drawn with ab as diameter

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### Mohammed

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## In a right triangle ABC in which = 90^o , 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 .

Click here👆to get an answer to your question ✍️ In a right triangle ABC in which = 90^o , 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 .

In a right triangle ABC in which ∠=90

Question o

, 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.

PQ and BQ are tangents drawn from an external point Q.

Medium Open in App Solution Verified by Toppr

PQ=BQ...................(i)     [Length of tangents drawn from an external point to the circle are equal]

∠PBQ=∠BPQ      [In atriangle, equal sides have equal angles opposite to them]

As, it is given that

AB is the diameter of the circle

∴∠APB=90 o

[Angle in a semi-circle is right angle]

∠APB+∠BPC=180 o [Linear pair] ∠BPC=180−90=90 o In △BPC ∠BPC+∠PBC+∠PCB=180 o

[Angle sum property]

∠PBC+∠PCB=180−90=90 o

..................(ii)

Now, ∠BPC=90 o ∠BPQ+∠CPQ=90 o ..............(iii)

From (ii) and (iii), we get

∠PBC+∠PCB=∠BPQ+∠CPQ

∠PCQ=∠CPQ        [∵∠BPQ=∠PBQ] [∠PCB=∠PCQ,∠PBQ=∠PBC]

In △PQC ∠PCQ=∠CPQ

∴PQ=QC................(iv)

From (i) and (iv), we get

BQ=QC

Thus, tangent at P bisects the side BC.

897 110

स्रोत : www.toppr.com

## 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

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

## 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 10NCERT 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 . . . .

स्रोत : www.cuemath.com

## 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.

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.

Byju's Answer In a right tr... Question

In a right triangle \$ ABC\$ in which \$ \angle 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\$.

(k. Open in App Solution

Step 1: Find the relation between the angles.

As we know, the angle in a semicircle is

90° . Therefore, ∠APB=90°

With the help of the property of the linear pair we get

∠BPC=90° . Therefore, ∠3+∠4=90° ...(1) .

Since, it is given that

∠ABC=90° . Therefore, ∠ABC+∠1+∠5=180° ⇒ ∠1+∠5=90° ...(2)

{Angle sum property of a triangle}

By alternate segment theorem we have:

∠1=∠3 So, the equation (2) becomes: ∠3+∠5=90° ...(3)

Step 2: Prove

PQ

bisects

BC

.

Subtract equation (1) from equation (3) . ∠3+∠5-∠3-∠4=0 ⇒ ∠5-∠4=0 ⇒ ∠5=∠4

Since, the sides opposite to equal angles are equal. Therefore,

PQ=QC ...(4)

Since, the tangent drawn to a circle from an external point are equal.

Therefore, BQ=PQ ...(5)

So, from the equation

(4) and equation (5) we get. BQ=QC

Hence it is proved that

PQ

bisects

BC

.

Suggest Corrections 20

SIMILAR QUESTIONS

Q. In a right triangle ABC in which

∠ B = 90 o

, 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.

Q. Question 6

In a right angle ∠ ABC is which ∠ B = 90 ∘

, 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.

Q. In a right triangle ABC in which B = 90, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Then the tangent to the circle at P bisects BC.Q. In a right triangle

A B C in which ∠ B = 90 ∘ ,

a circle is drawn with

A B

as diameter intersecting the hypotenuse

A C at P

then tangent to the circle at

P bisects B C.

Q. In a right triangle ABC in which\angle B = 90º, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the †an gent to the circle at P bisect BC.

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स्रोत : byjus.com

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Mohammed 3 day ago

Guys, does anyone know the answer?