Mathematics Part ii Solutions Solutions for Class 10 Math Chapter 2 Tangents are provided here with simple step-by-step explanations. These solutions for Tangents are extremely popular among class 10 students for Math Tangents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No 152:

#### Question 1:

Circles centred at A and B cut at P. Prove that if AP is a tangent to the circle centred at B, then BP is tangent to the circle centred at A.

#### Answer:

Given: Two circles with centre *A* and *B,* which cut each other at point *P.*

*AP* is a tangent to the circle centred at *B*.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

*AP *is a tangent to the circle centred at *B* and *BP* is the radius of the circle.

∴ ∠*BPA *= 90°

*BT *is a straight line passing through point *P*.

We know that sum of angles forming a linear pair is 180°.

∴ ∠*BPA *+ ∠*APT *= 180°

⇒ 90° + ∠*APT *= 180°

⇒ ∠*APT *= 90°

Therefore, *AP* is perpendicular to *BP*.

We know that a line drawn through any point of a circle, perpendicular to the radius through that point is a tangent to the circle.

Thus, *BP* is a tangent to the circle centred at *A*.

#### Page No 153:

#### Question 1:

Draw the figures below according to the specifications given.

#### Answer:

**(1)**

The steps of construction to construct the required figure are as follows:

1) Draw a circle with centre *O* and radius* OA* = 2 cm.

2) Extend line segment *OA* to a point *P* such that *OP* = 6 cm (= 4 cm + 2 cm).

3) Mark a point *B* on line segment *OP *such that *PB* = *BO* = = 3 cm.

4) Draw a circle with *B *as the centre and *BO* as the radius. Name the points where it intersects the previous circle at *S* and *T*.

5) Join *PS* and *PT*.

**(2)**

The given figure is:

We know that any tangent to circle is perpendicular to the radius to the point of contact.

*PS* is the tangent to the given circle and* SO *is the radius of the circle.

∴ ∠*OSP** *= 90°

Applying Pythagoras theorem in Δ*OSP*:

*OP*^{2} = *PS*^{2 }+ *SO*^{2}

Now, we can construct the required figure. The steps of construction are as follows:

1) Draw a circle with centre *O* and radius* OA* = 2.5 cm.

2) Extend line segment *OA* to a point *P* such that *OP* = 6.5 cm.

3) Mark a point *B* on line segment *OP *such that *PB* = *BO* = = 3.25 cm.

4) Draw a circle with *B *as the centre and *BO* as the radius. Name the points where it intersects the previous circle at *S* and *T*.

5) Join *PS*, *PT *and *OS*.

**(3)**

We have joined the centre of the circle given in the figure with the point where the tangent intersects the circle.

The given figure is:

We know that any tangent to circle is perpendicular to the radius to the point of contact.

*PS* is the tangent to the given circle and* SO *is the radius of the circle.

∴ ∠*OSP* = 90°

Applying Pythagoras theorem in Δ*OSP*:

*OP*^{2} = *PS*^{2 }+ *SO*^{2}

⇒ *SO*^{2} = *OP*^{2} − *PS*^{2}

Now, we can construct the required figure. The steps of construction are as follows:

1) Draw a circle with centre *O* and radius* OA* = 3.32 cm.

2) Extend line segment *OA* to a point *P* such that *OP* = 5.2 cm.

3) Mark a point *B* on line segment *OP *such that *PB* = *BO* = = 2.6 cm.

4) Draw a circle with *B *as the centre and *BO* as the radius. Name the points where it intersects the previous circle at *S* and *T*.

5) Join *PS* and *PT*.

#### Page No 153:

#### Question 2:

In the picture below, the radius of the circle is 15 centimetres. Compute the lengths of the tangents PQ and PR

(i) The point P is equidistant from A and B

(ii) The line OP bisects the line AB and the angle APB

#### Answer:

Given: *OR* = *OQ *= 15 cm and ∠*ROQ* = 120°

**Construction: **Join *OP*.

In Δ*OPR* and Δ*OPQ*:

*OR* = *OQ* (Radii of the same circle)

*OP* = *OP * (Common side)

*PR *= *PQ* (Length of tangents drawn from a point outside the circle are equal)

If all the sides of one triangle are equal to the corresponding sides of the other triangle then the two triangles are congruent.

∴ Δ*OPR* ≅ Δ*OPQ*

⇒ ∠*POR** *= ∠*POQ** * (Congruent parts of congruent triangles are congruent)

⇒ ∠*POR** *= ∠*POQ* = * *

In Δ*POR*:

= tan 60°

⇒ *RP* = *OR* tan 60°

= 15 cm ×

= 15 × 1.732 cm

= 25.98 cm

We know that lengths of tangents drawn from an external point are equal.

∴ *PQ* = *PR* = 25.98 cm

#### Page No 154:

#### Question 1:

In the circle centred at O, the tangents at A and B intersect at P. Prove the following:

(i)

the point P is equidistant from A and B

(ii)

the line OP bisects the line AB and the angle APB

(iii)

if the line OP cuts the line AB at Q, then OQ × OP = *r*^{2}, where *r* is the radius of the circle

#### Answer:

Given: A circle with centre *O*

Tangents at points *A* and *B *intersect at point *P*.

(i)

**Construction:** Join *AO*, *OB* and *OP*.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

*AP *is a tangent to the circle and *OA* is the radius of the circle.

∴∠*OAP* = 90°

Applying Pythagoras theorem in Δ*OAP*:

*OP*^{2} = *OA*^{2} + *AP*^{2}

_{⇒}*OA*^{2} = *OP*^{2} − *AP*^{2} … (1)

Similarly, we have *OB*^{2} = *OP*^{2} − *BP*^{2}^{ }* * …(2)

*OA* = *OB* (Radii of the same circle)

From equations (1) and (2), we have:

*OP*^{2} − *AP*^{2} = *OP*^{2} − *BP*^{2}^{ }* *

⇒ *AP*^{2} = *BP*^{2}

⇒ *AP*^{ }= *BP*

Thus, point *P* is equidistant from *A* and *B*.

(ii)

**Construction:** Join *AB*.

In Δ*OAP* and Δ*OBP*:

*PA* = *PB * (Proved in part (1))

*PO* = *PO* (Common side)

*OA* = *OB * (Radii of the same circle)

As all sides of one triangle are equal to the corresponding sides of the other triangle, Δ*OAP *≅ Δ*OBP*.

Corresponding parts of congruent triangles are congruent.

⇒∠*AOP* = ∠*BOP** *…(1)

∠*APO* = ∠*BPO *

Thus, line segment *OP* bisects ∠*APB*.

In Δ*AQO* and Δ*BQO*:

*OA* = *OB * (Radii of the same circle)

∠*AOP* = ∠*BOP** * (From equation (1))

*OQ* = *OQ * (Common side)

As two sides and included angle of one triangle are equal to the two sides and included angle of the other triangle, Δ*AQO *≅ Δ*BQO*.

Corresponding parts of congruent triangles are congruent.

⇒ *AQ* = *BQ*

Thus, line segment* OP *bisects *AB*.

(iii)

We have shown in part (2) that Δ*AQO** *≅ Δ*BQO*.

∴ ∠*AQO* = ∠*BQO** * (Corresponding parts of congruent triangles are congruent)

We know that sum of angles forming a linear pair are equal.

∴ ∠*AQO* + ∠*BQO** *= 180°

⇒ ∠*AQO** *= ∠*BQO** *= 90°

In Δ*AQO* and Δ*PBO*:

∠*AQO** *= ∠*PBO* = 90°

∠*AOP* = ∠*BOP* (Proved in part (1))

∴ Δ*AQO** *∼ Δ*PBO* (By angle angle criterion)

#### Page No 154:

#### Question 2:

The circle in the figure below touches all the three lines.

Prove that the perimeter of the right angle triangle is equal to the diameter of the circle.

#### Answer:

**Construction:** Joining the centre *O* of the circle with the point at which the tangents touch the circle.

After naming, we have the following figure.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

*AE *is the tangent to the circle and *OE* is the radius of the circle.

∴ ∠*OEA* = 90°

Similarly, ∠*AFO* = 90°

In quadrilateral *AFOE*:

∠*A* = ∠*OEA* = ∠*OFA* = 90°

Thus, by angle sum property of quadrilaterals:

∠*EOF* = 90°

Also, *OE* = *OF* = *r* (say)* * (Radii of the same circle)

As all angles of quadrilateral *OEAF* are of measure 90° and adjacent sides are equal, quadrilateral *OEAF* is a square.

We know that lengths of tangents drawn from a point outside a circle are equal.

⇒ *AE *= *AF *

⇒ *AE *= *AF *= *OE* = *OF* = *r* (All sides of a square are equal in length)

Now, *AE* + *AF* = *r *+ *r*

= 2*r*

= *d* (where *d* is the diametre of the circle) …. (1)

Also, *DE* =* DC *and* BC *=* BF *(Lengths of tangents drawn from a point outside a circle) …(2)

We know that perimeter of a triangle is equal to the sum of the lengths of its sides.

∴ Perimeter of Δ*DAB* = *AD *+ *DB* + *BA*

= *AD* + *DC* +* CB *+ *BA*

= *AD *+ *DE *+ *BF *+* BA *(From equation (2))

=* AE *+ *AF *

= *d * (From equation (1))

Therefore, the perimeter of the right-angled triangle is equal to the diametre of the circle.

#### Page No 154:

#### Question 3:

In the figure below, the lines AB, BC, CA touch the circle at P, Q, R.

Prove that the perimeter of the triangle is 2(AP + BQ + CR)

#### Answer:

We know that perimeter of a triangle is equal to the sum of the lengths of its sides.

Perimeter of Δ*ABC* =* AB* + *BC* +* CA*

= (*AP* + *PB*) + (*BQ* + *QC*) + (*AR* + *RC*) …(1)

We know that lengths of tangents drawn from a point outside a circle are equal.

⇒ *AP *= *AR* ... (2)

*BP *= *BQ * ... (3)

*CR* = *CQ * ... (4)

Substituting the values from equations (2), (3) and (4) in equation (1):

Perimeter of Δ*ABC* = *AP* + *BQ* + *BQ *+* CR *+ *AP* +* CR*

= 2(*AP* + *BQ* + *CR*)

#### Page No 161:

#### Question 1:

Draw a circle of radius 3 centimetres and draw a rhombus with one angle 40°, all four sides touching the circle.

#### Answer:

The rough sketch of the required figure can be drawn as follows:

We know that the central angle of the smaller arc between the two points on a circle and the angle between the tangents at these points are supplementary.

∠*ECF* + ∠*EOF* = 180°

⇒ ∠*EOF* = 180° − 40° = 140°

We also know that any tangent to circle is perpendicular to the radius to the point of contact.

∴ ∠*OFC* = ∠*OGB* = ∠*OHD* = ∠*OEC* = 90°

We will use these measurements to construct the required figure.

The steps of construction are as follows:

1) Draw a circle with centre *O* and radius, *OF *= 3 cm.

2) Draw ∠*OFX* of measure 90° at point* F* by taking *OF* as the base. Extend ray* XF* downwards.

3) Draw ∠*EOF* of measure 140° at point *O* by taking *FO* as the base.

4) Draw ∠*OEC* of measure 90° at point *E* by taking *OE* as the base. Extend *CE* downwards.

5) Extend the line segments *EO* and *OF* such that they intersect the circle at points *G* and H respectively.

6) Draw ∠*OGB* of measure 90° at point *G* by taking* OG* as the base, where *B* is a point on line *FX*. Extend *BG* downwards.

7) Draw ∠*OHD* of measure 90° at point *H* by taking *OH* as the base, where *D* is a point on ray *CE*.

8) Extend line segment* DH* to intersect ray *BG* at point *A*.

*ABCD* is the required rhombus with all the four sides touching the circle.

#### Page No 161:

#### Question 2:

Draw a circle of radius 4 centimetres and draw a regular pentagon with all its sides touching the circle.

#### Answer:

In a polygon, sum of all the sides is given by (*n* − 2) × 180°, where *n *is the number of sides.

For a regular pentagon, *n* = 5

Sum of all the angles = (5 − 2) × 180°

= 3 × 180°

= 540°

Since all the angles of a regular pentagon are equal, measure of each angle = = 108°.

The rough sketch of the required figure can be drawn as follows:

We know that the central angle of the smaller arc between the two points on a circle and the angle between the tangents at these points are supplementary.

∴ ∠*PBQ* + ∠*POQ* = 180°

⇒ 108° + ∠*POQ* = 180°

⇒ ∠*POQ* = 180° − 108° = 72°

We also know that any tangent to circle is perpendicular to the radius to the point of contact.

∴ ∠*OPB* = ∠*OQB *= ∠*OTA* = ∠*OSE *= ∠*ORD *= 90°

We will use these measurements to construct the required figure.

The steps of construction are as follows:

1) Draw a circle with centre *O* and radius, *OQ *= 4 cm.

2) Draw ∠*OQX* of measure 90° at point* Q* by taking *OQ *as the base. Extend ray* QX *upwards.

3) Draw ∠*QOP* of measure 72° at point *O* by taking *QO* as the base such that point *P* lies on the circle.

4) Draw ∠*OPB* of measure 90° at point *P *by taking *OP* as the base such that* B *is a point on line *QX*. Extend ray *PB* on the left side.

5) Draw ∠*POT* of measure 72° at point *O* by taking *PO* as the base such that point *T* lies on the circle.

6) Draw ∠*OTA* of measure 90° at point *T *by taking *OT* as the base such that *A* is a point on line *PB*. Extend ray *TA* upwards.

7) Draw ∠*TOS* of measure 72° at point *O* by taking *TO* as the base such that point *S *lies on the circle.

8) Draw ∠*OSE* of measure 90° at point *S *by taking *OS* as the base such that *E* is a point on line *TA*. Extend ray *ES* upwards.

9) Draw ∠*SOR *of measure 72° at point *O* by taking *SO *as the base such that point *R *lies on the circle.

10) Draw ∠*ORD* of measure 90° at point *R *by taking *OR* as the base such that *D* is a point on line *SE*.

11) Extend ray* RD *downwards to intersect line *QX* at point *C*.

*ABCDE* is the required pentagon with all the sides touching the circle.

#### Page No 161:

#### Question 3:

Prove that in any circle, the tangents at two points make equal angles with the chord joining the points of contact.

#### Answer:

Given: A circle with centre *O*

*PA* and *PB* are tangents to the circle through a point *P *and *AB* is the chord.

We know that lengths of tangents drawn from a point outside the circle are equal.

∴ *PA* = *PB*

In Δ*PAB*, *PA* = *PB*

We know that angles opposite to equal sides are equal in measure.

∴ ∠*PBA** *= ∠*PAB*

Thus, in a circle, the tangents at two points make equal angles with the chord joining the points of contact.

#### Page No 161:

#### Question 4:

In the figure below, all the vertices of the small triangle are on the circle and all the sides of the larger triangle touch the circle at these points.

Find all angles of the small triangle.

#### Answer:

Using angle sum property in Δ*ABC*:

∠*ABC** *+ ∠*BCA* + ∠*CAB** *= 180°

⇒ 30° + 70° + ∠*CAB* = 180°

⇒ 100° + ∠*CAB* = 180°

⇒ ∠*CAB** *= 180° −100° = 80°

We know that lengths of tangents drawn from a point outside the circle are equal.

∴ *BQ* = *BR*

*AQ* = *AP*

*CP *= *CR*

In Δ*BQR*, *BQ* = *BR*

We know that angles opposite to equal sides are equal in measure.

∠*BRQ* = ∠*BQR*

By angle sum property:

∠*BQR* + ∠*QRB* + ∠*RBQ* = 180°

⇒ 2∠*BQR** *+ 30° = 180°

⇒ 2∠*BQR* = 150°

⇒ ∠*BQR* = ∠*BRQ* = 75°

Similarly, in Δ*RPC*:

∠*CRP* = ∠*CPR* (As *CP *= *CR*)

By angle sum property:

∠*CRP* + ∠*RPC* + ∠*PCR** *= 180°

⇒ 2∠*CRP** *+ 70° = 180°

⇒ 2∠*CRP** *= 110°

⇒ ∠*CRP* = ∠*CPR* = 55°

In Δ*AQP*:

∠*APQ* = ∠*AQP** * (As *AQ* = *AP*)

By angle sum property:

∠*APQ* + ∠*PQA* + ∠*QAP* = 180°

⇒ 2∠*APQ* + 80° = 180°

⇒ 2∠*APQ** *= 100°

⇒ ∠*AQP* =∠*APQ* = 50°

We know that each angle between a chord and the tangent at one of its ends in a circle is equal to the angle in the segment on the other side of the chord.

∴ ∠*BQR* = ∠*RPQ* = 75°

∠*RPC* = ∠*RQP* = 55°

∠*AQP* = ∠*QRP** *= 50°

Hence, the measures of all the angles of smaller triangle are 75°, 55° and 50°.

#### Page No 161:

#### Question 5:

In the picture below, PQ, RS, TU are tangents to the circle at A, B, C.

How many pairs of equal angles are there in it?

#### Answer:

In a circle, each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord.

So, the pair of equal angles are:

∠*BAC* = ∠*UCB* …(1)

∠*BAC* = ∠*SBC** * …(2)

∠*ABC* = ∠*TCA** * …(3)

∠*ABC** *= ∠*PAC** * …(4)

∠*ACB** *= ∠*QAB* …(5)

∠*ACB* = ∠*RBA* …(6)

From (1) and (2):

∠*UCB** *= ∠*SBC*

From (3) and (4):

∠*TCA* = ∠*PAC** *

From (5) and (6):

∠*QAB *= ∠*RBA** *

So, there are nine pairs of equal angles in the given figure.

#### Page No 164:

#### Question 1:

Draw a triangle of sides 4, 5, 6 centimetres and draw its incircle.

#### Answer:

The incircle for triangle with sides 4 cm, 5 cm and 6 cm as can be constructed as follows:

1) Draw a line segment *AB* of length 5 cm.

2) Taking *A* as the centre and 5 cm as the radius, draw an arc on the upper side of *AB*.

3) Taking *B* as the centre and 6 cm as the radius, draw an arc cutting the previously drawn arc at point *C*.

4) Join* AC* and *BC*.

Δ*ABC* is the required triangle.

The incircle for the given equilateral triangle can be constructed as follows:

1) Draw angle bisectors of ∠*A* and ∠*B*. Let them intersect at point *O*.

2) With *O *as the centre, draw* OK *perpendicular to *AC*. Name the point where it intersects *AC* as *D*.

3) With *O* as the centre and *OD* as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given triangle.

#### Page No 164:

#### Question 2:

Draw an equilateral triangle of sides 6 centimetres and draw its incircle and circumcircle.

#### Answer:

An equilateral triangle with side 6 cm can be constructed as follows:

1) Draw a line segment *AB* of length 6 cm.

2) Taking *A* as the centre and 6 cm as the radius, draw an arc on the upper side of *AB*.

3) Taking* B *as the centre and 6 cm as the radius, draw an arc cutting the previously drawn arc at point *C*.

4) Join *AC* and* BC*.

Δ*ABC* is the required equilateral triangle.

The incircle for the given equilateral triangle can be constructed as follows:

1) Draw angle bisectors of ∠*A* and ∠*B*. Let them intersect at point *O*.

2) With *O* as the centre, draw *OK* perpendicular to *AC*. Name the point where it intersects *AC* as *D*.

3) With *O* as the centre and *OD* as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given triangle.

The circumcircle for the given equilateral triangle can be constructed as follows:

1) Draw perpendicular bisectors of *AB* and *AC*. Let these perpendicular bisectors intersect at point *O*.

2) With *O *as the centre and radius *OA *=* OB *= *OC*, draw a circle passing through the points *A*, *B* and *C*.

The given figure shows the required circumcircle of the given triangle.

#### Page No 164:

#### Question 4:

Draw a square of sides 5 centimetres and draw its circumcircle and incircle.

#### Answer:

A square with side 5 cm as length can be drawn as follows:

1) Draw a line segment *AB* of length 5 cm.

2) At point* A *and *B*, draw ∠*XAB* = ∠*YBA* = 90°.

3) Taking *A* and *B* as the centres and 5 cm as the radius, draw arcs on rays* AX *and *BY*. Name the point of intersection as *D* and *C*.

4) Join *DC*.

Quadrilateral *ABCD* is the required square.

The incircle of this square can be drawn as follows:

1) Draw angle bisectors of ∠*A* and ∠*B*. Let them intersect at point *O*.

2) With* O *as the centre, draw *OK* perpendicular to *AB*. Name the point where it intersects *AB* as *P*.

3) With* O *as centre and *OP* as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given square.

The circumcircle for the given square can be constructed as follows:

1) Draw perpendicular bisectors of* AB* and *BC*. Let them intersect at point *O*.

2) With* O* as the centre and radius *OA* = *OB* = *OC* = *OD*, draw a circle passing through the points *A*, *B*, *C* and *D*.

The given figure shows the required circumcircle of the given square.

#### Page No 164:

#### Question 5:

Draw the figure below according to the given specifications:

#### Answer:

The given figure is:

We know that sum of interior angles on the same side of the transversal are supplementary.

∴∠*A* + ∠*B* = 180°

⇒ 50° + ∠*B* = 180°

⇒ ∠*B* = 180° − 50° = 130°

The given rhombus can be drawn as follows:

1) Draw a line segment *AB* of length 5 cm.

2) At point *A*, draw ∠*TAB* = 50°.

3) Taking *A* as the centre and 5 cm as the radius, draw an arc on ray *AT*. Name the point of intersection as *D*.

4) At point *B*, draw ∠*UBA* = 130°.

5) Taking *B* as the centre and 5 cm as the radius, draw an arc on ray *BU.* Name the point of intersection as *C*.

6) Join *CD*.

Quadrilateral *ABCD* is the required rhombus.

The incircle of this rhombus can be made as follows:

1) Draw angle bisectors of ∠*A* and ∠*B*. Let them intersect at point* O*.

2) With *O* as the centre, draw *OK* perpendicular to *AB*. Name the point where it intersects *AB* as* P*.

3) With *O* as centre and *OP* as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given rhombus.

#### Page No 164:

#### Question 3:

Prove that in an equilateral triangle, the circumcentre and incentre are the same. What is the ratio of the circumradius and inradius?

#### Answer:

**Case I.**

Consider an equilateral triangle *ABC*.

Let *O*′ be its incentre.

We know that incentre of a circle is obtained by the intersection of the angles bisectors of a triangle.

*BO*′ is the angle bisector of ∠*ABC*.

∴ ∠*ABO*′ = ∠*O*′*BC** *=

Similarly, we have ∠*O*′*CA** *= ∠*O*′*AC* = ∠*O*′*AB* = ∠*O*′*BA* = 30°

**Case II.**

Now, again consider the same triangle with angles marked on it.

Let *O* be the circumcentre of the given triangle.

As *O* is the circumcentre of Δ*ABC*,* OA *= *OB *= *OC*

In Δ*OBC*, *OB *= *OC*

⇒ ∠2 = ∠1 …(1)

Also, ∠1 + ∠6 = ∠2 + ∠3 (As each angle of an equilateral triangle is equal)

⇒ ∠6 = ∠3 …(2)

Similarly, in Δ*OAB*:

∠5 = ∠6 …(3)

In Δ*OAC*:

∠3 = ∠4 …(4)

From equations (2) and (4), we have:

∠4 = ∠6 …(5)

From equations (3) and (5), we have:

∠4 = ∠5

As ∠4 + ∠5 = 60°

⇒ 2∠4 = 60°

⇒ ∠4 = 30°

Similarly, we can show that ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = 30°, i.e., ∠*ABO* = ∠*OBC* = ∠*OCA** *= ∠*OAC* = ∠*OAB** *= ∠*OBA* = 30°.

Therefore,* BO*, *AO* and *CO* is the angle bisectors of the angles of the given triangle *ABC*.

However, *BO*′, *AO*′ and *CO*′ is the angle bisectors of the angles of the given triangle *ABC* according to our assumption in case I.

Thus, *O* and *O*′ coincide.

Hence, the circumcentre and incentre of an equilateral triangle are same.

We know that centroid of an equilateral triangle coincides with circumcentre and orthocentre.

Also, centroid divides the median in the ratio 2:1.

Here, *AD* is the median.

⇒ *AO* = and *OD *=

⇒ *AD* = 3 *OD*

⇒ *AO *= 2 *OD*

⇒ *AO* : *OD* = 2 : 1

Thus, the ratio of the circumradius to inradius is 2:1.

View NCERT Solutions for all chapters of Class 10