SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024

### Triangles Class 9th Mathematics Part Ii MHB Solution

##### Triangles Class 9th Mathematics Part Ii MHB Solution

###### Practice Set 3.1
Question 1.

In figure 3.8, ∠ACD is an exterior angle of Î”ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.

Given, ∠A = 70° and ∠B = 40°

In a triangle ABC,

The measure of an exterior angle of a triangle is equal to the sum of its remote interior angles

∠ACD is an exterior angle of triangle ABC

So, from theorem of remote interior angles,

∠ACD = ∠BAC + ∠ABC

⇒ ∠ACD = ∠A + ∠B

⇒ ∠ACD = 70° + 40° = 110°

Question 2.

In Î”PQR, ∠P = 70° ∠Q = 65° then find ∠R.

Given, ∠P = 70° ∠Q = 65°

In a triangle we know sum of interior angles is 180°

∴ in Î”PQR

∠P + ∠Q + ∠R = 180°

70° + 65° + ∠R = 180°

∠R = 180° - 135° = 45°

Question 3.

The measures of angles of a triangle are x°, (x – 20)°, (x – 40)°. Find the measure of each angle.

Angles of a triangle are

In a triangle we know sum of interior angles is 180°

∴ x° + (x – 20)° + (x – 40)° = 180°

x° + x° - 20° + x° - 40° = 180°

3x° = 180° + 60°

x° = 240°/3

∴ x° = 80°

Angles of the triangle are x° = 80°

(x – 20)° = 80° - 20° = 60°

(x – 40)° = 80° - 40° = 40°

Question 4.

The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.

Let the measure of the smallest angle be x

Measure of second angle = 2x

Measure of third angle = 3x

In a triangle we know sum of interior angles is 180°

∴ x + 2x + 3x = 180°

⇒ 6x = 180°

⇒ x = 180°/6

⇒ x = 30°

Measure of smallest angle = x = 30°

Measure of second angle = 2x = 2 × 30° = 60°

Measure of third angle = 3x = 3 × 30° = 90°

Question 5.

In figure 3.9, measures of some angles are given. Using the measures find the values of xyz.

Given ∠TEN = 100°, ∠EMR = 140°

∠NEM = y, ∠ENM = x, ∠NME = z

In a triangle ENM

The measure of an exterior angle of a triangle is equal to the sum of its remote interior angles

∠TEN and ∠EMR is an exterior angle of triangle ENM

So from theorem of remote interior angles,

∠TEN = ∠ NME + ∠ENM

⇒ 100° = z + x ……. (1)

∠EMR = ∠NEM + ∠ENM

⇒ 140° = x + y

⇒ x = 140° - y …(2)

In a triangle we know sum of interior angles is 180°

∴ x + y + z = 180 ……….(3)

Putting (1) in (3)

⇒ y + 100° = 180°

⇒ y = 180° - 100° = 80°

Putting y in (2)

∴ x = 140° - 80°

⇒ x = 60°

Putting x in (1)

∴ 60° + z = 100°

⇒ z = 100° - 60°

⇒ z = 40°

Measure of all the angles are

x = 60°, y = 80°, z = 40°

Question 6.

In figure 3.10, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

Given ∠DAB = 70° and ∠DER = 40°

In the given figure ∠DAB = ∠ADE [Alternate Interior angles are equal]

∴ ∠ADE = ∠RDE = 70°

In Î”DER,

∠DER + ∠DRE + ∠RDE = 180°

⇒ 40° + ∠DRE + 70° = 180°

⇒ ∠DRE = 180° - 110°

⇒ ∠DRE = 70°

∵ ∠ARE is an exterior angle of triangle DER

∠ARE = ∠RDE + ∠DER = 70° + 40°

⇒ ∠ARE = 110°

Question 7.

In Î”ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C= 700 Find measure of ∠AOB.

The figure is attached below:

BN and AM are the angle bisectors of angle B and A respectively.

Given ∠C = 70°

In a triangle we know sum of interior angles is 180°

In Î”ABC

∠A + ∠B + ∠C = 180°

∠A + ∠B = 180° - 70°

∠A + ∠B = 110°

Now in Î”AOB

AO is the bisector of ∠A

BO is the bisector of ∠B

∴ ∠OAB = ∠A/2 and ∠OBA = ∠B/2

∠OAB + ∠OBA + ∠AOB = 180°

∠A/2 + ∠B/2 + ∠AOB = 180°

⇒ ∠AOB = 180° - (∠A + ∠B)/2

⇒ ∠AOB = 180° - 110°/2 = 180° - 55°

⇒ ∠AOB = 125°

Question 8.

In Figure 3.11, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively.

Prove that ∠PTQ = 90°.

Given: AB || CD, line PQ is the tranversal

Ray PT and Ray QT are bisectors of ∠BPQ and ∠PQD

To prove: ∠PTQ = 90°

Proof: Since, Ray PT and Ray QT are bisectors of ∠BPQ and ∠PQD

∠TPQ = ∠BPQ/2 ……..(1)

∠PQT = ∠PQD/2 ………(2)

Since, two parallel lines are intersected by a transversal, the interior angles on either side of the transversal are supplementary.

So, ∠BPQ + ∠PQD = 180°

Dividing both sides by 2, we get

⇒ (∠BPQ + ∠PQD)/2 = 180°/2

⇒ ∠BPQ/2 + ∠PQD/2 = 90°

In Î”PQT,

∠TPQ + ∠PQT + ∠PTQ = 180°

Substituting ∠TPQ and ∠PQT from (1) and (2) respectively

⇒ ∠BPQ/2 + ∠PQD/2 + ∠PQT = 180°

⇒ 90° + ∠PQT = 180°

⇒ ∠PQT = 180° - 90°

⇒ ∠PQT = 90°

Hence, proved.

Question 9.

Using the information in figure 3.12, find the measures of ∠a, ∠b and ∠c.

In the given triangle

a + b + c = 180° …………(1)

c + 100° = 180° ……….(2) [angles in linear pair]

⇒ c = 180° - 100°

⇒ c = 80°

b = 70° ……………..(3) [opposite angles are equal]

Putting value of b and c in (1)

⇒ a + 70° + 80° = 180°

⇒ a = 180° - 150°

⇒ a = 30°

Question 10.

In figure 3.13, line DE || line GF ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively.

Prove that,

i. ∠DEG = 1/2∠EDF

ii. EF =FG.

Given: line DE || line GF

Ray EG and ray FG are bisectors of  and  respectively

To Prove: i.

ii.

Proof: Ray EG and ray FG are bisectors of  and  respectively.

So, ∠DEG = ∠GEF = 1/2 ∠DEF ……………..(1)

∠DFG = ∠GFM = 1/2 ∠DFM ………..(2)

Also, ∠EDF = ∠DFG …..(3) [Alternate interior angles]

In Î”DEF

∠DFM = ∠DEF + ∠EDF

From (2) and (3)

2∠EDF = ∠DEF + ∠EDF

⇒ ∠EDF = ∠DEF

From (1)

⇒ ∠EDF = 2∠DEG

⇒ ∠DEG = 1/2 ∠EDF

Hence, (i) is proved.

Line DE || line GF

From alternate interior angles

∠DEG = ∠EGF …….(4)

From (1)

∠GEF = ∠EGF

Since, in the Î”EGF sides opposite to equal angles are equal.

∴ EF = FG

Hence, (ii) is proved.

###### Practice Set 3.2
Question 1.

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

: By SSS congruency test

Î”ABC ≅ Î”PQR

Explanation:

Given, AB = PQ

BC = QR

CA = RP

∴ By SSS congruency test

Î”ABC ≅ Î”PQR

SSS : Side Side Side

Question 2.

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

By SAS congruency test

Î”XYZ ≅ Î”LMN

Explanation:

Given: XY = LM

∠XYZ = ∠LMN

YZ = MN

Therefore, By SAS congruency test

Î”XYZ ≅ Î”LMN

SAS: Side Angle Side

Question 3.

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

By ASA congruency test

Î”PRQ ≅ Î”STU

Explanation:

Given: ∠QPR = ∠UST

PR = ST

∠PRQ = ∠STU

Therefore, By ASA congruency test

Î”PRQ ≅ Î”STU

ASA: Angle Side Angle

Question 4.

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

By RHS congruency test

Î”LMN ≅ Î”PTR

Explanation:

Given: LM = PT

∠LMN = ∠PTR

LN = PR

Therefore, By RHS congruency test

Î”LMN ≅ Î”PTR

RHS: Right Hypotenuse Side

Question 5.

Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

From the information shown in the figure, in Î”ABC and Î”PQR

∠ABC ≅ ∠PQR

seg BC ≅ seg QR

∠ABC = ∠PRQ

∴ Î”ABC ≅ Î”PQR ……………………____ test

∴ ∠BAC = _____……………………corresponding angles of congruent triangles.

seg AB ≅ ____………………….. corresponding sides of congruent triangles.

____ = seg PR …………………corresponding side of congruent triangles.

Given: ∠ABC = ∠PQR

BC = QR

∠ABC = ∠PRQ

∴ Î”ABC ≅ Î”PQR …………………………ASA test

ASA: angle side angle

∴ ∠BAC = ∠QPR …………………….corresponding angles of congruent triangles.

seg AB = seg PQ ………………….. corresponding sides of congruent triangles.

seg AC = seg PR ………………….. corresponding angles of congruent triangles.

Question 6.

Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

From the information shown in the figure.,

In Î”PTQ and Î”STR

∠PTQ = ∠STR ……………………….. vertically opposite angles

seg TQ ≅ seg TR

∴ Î”PTQ ≅ Î”STR ………………._____ test

∠TPQ ≅ ____ …………………… corresponding angles of congruent triangles.

_____ ≅ ∠TRS ………………….. corresponding angles of congruent triangles.

seg PQ ≅ _____ …………….. corresponding sides of congruent triangles.

In Î”PTQ and Î”STR

Given: ∠PTQ = ∠STR ……………………….. vertically opposite angles

seg TQ = seg TR

seg TP = seg TS

∴ Î”PTQ = Î”STR ……………….SAS test

SAS: side angle side

∠TPQ = ∠TSR …………………… corresponding angles of congruent triangles.

∠TQP = ∠TRS ………………….. corresponding angles of congruent triangles.

seg PQ = seg SR …………….. corresponding sides of congruent triangles.

Question 7.

From the information shown in the figure, state the test assuring the congruence of Î”ABC and Î”PQR Write the remaining congruent parts of the triangles.

In Î”ABC and Î”PQR

AB = QP

BC = PR

∠CAB = ∠RQP

∴ By RHS congruency test

Î”ABC ≅ Î”QPR

∴ AC = QR ……………….. corresponding sides of congruent triangles.

∠ABC = ∠ QPR ………………… corresponding angles of congruent triangles.

∠BCA = ∠PRQ ………………… corresponding angles of congruent triangles.

Question 8.

As shown in the following figure, in Î”LMN and Î”PMN, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts

Given, In Î”LMN and Î”PNM

LM = PN

LN = PN

MN = MN

∴ By SSS congruency test

Î”LMN ≅ Î”PNM

∠LMN = ∠ PNM ………………… corresponding angles of congruent triangles.

∠LNM = ∠PMN ………………… corresponding angles of congruent triangles.

∠NLM = ∠MPN ……………….. corresponding angles of congruent triangles.

Question 9.

In figure 3.24, seg AB ≅ seg CB and seg AD ≅ seg CD.

Prove that Î”ABD ≅ Î”CBD

Given, In Î”ABD and Î”CBD

AB = CB

BD = BD ……….[Common]

∴ By SSS congruency test

Î”ABD ≅ Î”CBD

Question 10.

In figure 3.25, ∠P ≅ ∠R seg PQ ≅ seg RQ

Prove that, Î”PQT ≅ Î”PQS

In Î”PQT and Î”RQS

∠P = ∠R …………[Given]

∠QPT = ∠QRS

PQ = RQ ………….[Given]

∠PQT = ∠RQS ………….[common]

∴ By ASA congruency

Î”PQT ≅ Î”RQS

###### Practice Set 3.3
Question 1.

Find the values of x and y using the information shown in figure 3.37. Find the measure of ∠ABC and ∠ACB.

In Î”ABC

Given, AB = AC

Sides of a triangle are Equal then the angles opposite to them are equal.

∠ABC = ∠ACB

∴ x = 50°

So, ∠ABD = 50° + 60° = 110°

In Î”DBC

Given, DB = DC

Sides of a triangle are Equal then the angles opposite to them are equal.

∠DBC = ∠DCB

∴ y = 60°

∠ACD = ∠ACB + ∠BCD

= 50° + 60°

∴ ∠ACD = 110°

Question 2.

The length of hypotenuse of a right-angled triangle is 15. Find the length of median of its hypotenuse.

Length of hypotenuse of right-angled triangle = 15

We know, the length of the median of the hypotenuse is half the length

of the hypotenuse.

i.e.

Length of median of its hypotenuse = 1/2 × length of hypotenuse

Length of median of its hypotenuse = 1/2 × 15

= 7.5

∴ Length of median of its hypotenuse is 7.5

Question 3.

In Î”PQR, ∠Q =900, PQ = 12, QR = 5 and QS is a median. Find t(QS).

Î”PQR is a right-angled triangle

So, PQ and QR are the sides and PR is the hypotenuse of Î”PQR.

∴ By Pythagoras theorem

PQ2 + QR2 = PR2

⇒ PR2 = 122 + 52 = 144 + 25 = 169

⇒ PR = 13

Length of hypotenuse of right-angled triangle = 13

We know, the length of the median of the hypotenuse is half the length

of the hypotenuse.

i.e.

Length of median of its hypotenuse = 1/2 × length of hypotenuse

Length of median of its hypotenuse = 1/2 × 13

= 6.5

∴ Length of median of its hypotenuse is 6.5

Question 4.

In figure 3.38, point G is the point of concurrence of the medians of Î”PQR. If GT = 2.5, find the lengths of PG and PT.

Given, in Î”PQR

GT = 2.5

The point of concurrence of medians of a triangle divides each median in

the ratio 2 : 1.

Since, PT is the median.

∴ PG: GT = 2: 1

⇒ PG = 2 × 2.5 = 5

Therefore, length of PG = 5

Length of PT = PG + GT

= 5 + 2.5

Length of PT = 7.5

###### Practice Set 3.4
Question 1.

In figure 3.48, point A is on the bisector of ∠XYZ. If AX = 2cm then find AZ.

Given, Point A is on the bisector of ∠XYZ

AX = 2cm

Every point on the bisector of an angle is equidistant from the sides of the

angle.

Therefore, from figure

AX = AZ

∴ AZ = 2 cm

Question 2.

In figure 3.49, ∠RST = 56°, seg PT ⊥ ray ST, seg PT ⊥ ray ST, seg PR ⊥ ray SR and PR ≅ seg PT Find the measure of ∠RSP. State the reason for your answer.

Given, ∠RST = 56°

PT perpendicular to ST

PR perpendicular to SR

PR ≅ PT

Since, PR ≅ PT

∴ Any point equidistant from sides of an angle is on the bisector of the

angle.

Therefore, Ray SP is the bisector of ∠TSR.

That is ∠RSP = ∠TSP

Now, ∠RST = ∠RSP + ∠TSP

= 2 ∠RSP

∠RSP = 1/2 ∠RST

∠RSP = 1/2 × 56°

Therefore, ∠RSP = 28°

Question 3.

In Î”PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.

Given, in Î”PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm

We know, If two sides of a triangle are unequal, then the angle opposite to

the greater side is greater than angle opposite to the smaller side.

Here greater side is PQ and the smallest side is PR

∴ Angle opposite to QR = ∠QPR

Angle opposite to PR = ∠PQR

Greatest angle of triangle = ∠QPR

Smallest angle of triangle = ∠PQR

Question 4.

In Î”FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.

Given In Î”FAN,

∠F = 80°, ∠A = 40°

In a triangle sum of interior angles of the triangle is 180°

∴ ∠F + ∠A + ∠N = 180°

⇒ 80° + 40° + ∠N = 180°

⇒ ∠N = 180° - 120°

⇒ ∠N = 60°

So, ∠F = 80°, ∠N = 60°, ∠A = 40°

If two angles of a triangle are unequal then the side opposite to the greater.

Angle is greater than the side opposite to smaller angle.

Here greatest angle is ∠F and the smallest angle is ∠A

Side opposite to ∠F = NA

Side opposite to ∠A = FN

Greatest side of triangle = NA

Smallest side of triangle = FN

Question 5.

Prove that an equilateral triangle is equiangular

Given: Equilateral triangle PQR

To Prove: ∠P ≅ ∠Q ≅∠R

Proof: PQ ≅ PR ……….[all sides of an equilateral triangle are congruent.]

∠Q ≅ ∠R [the angles opposite to the two congruent sides of a triangle are congruent (Isosceles Triangle Theorem)]

PQ ≅ QR [since all sides of an equilateral triangle are congruent.]

∠R ≅ ∠P, again, by the Isosceles Triangle Theorem

Now, since ∠Q ≅ ∠R and ∠R ≅ ∠P ,

So, ∠Q ≅ ∠P

Therefore, ∠P ≅ ∠Q.

So, equilateral triangles are equiangular.

Question 6.

Prove that, if the bisector of ∠BAC of Î”ABC is perpendicular to side BC, then Î”ABC is an isosceles triangle.

Given: Bisector of ∠BAC of Î”ABC is perpendicular to side BC

To Prove: Î”ABC is an isosceles triangle.

Proof:

In Î”ABD and Î”ACD

Since, AD is the angle Bisector of Î”ABC

So, by ASA congruency test

Î”ABD ≅ Î”ACD

Therefore,

AB = AC ………………. corresponding sides of congruent triangles.

∠ABD = ∠ACD ……………… corresponding angles of congruent triangles.

∴ ∠ABC = ∠ACB

Since, AB = AC and ∠ABC = ∠ACB so, Î”ABC is an isosceles triangle.

Question 7.

In figure 3.50, if seg PR ≅ seg PQ, show that seg PS > seg PQ.

Given:

To prove:

Proof:

In Î”PRQ

PQ = PR …………….[given]

∠R = ∠PQ ....(i) [Angles opposite to equal sides are equal]

∠PQR > ∠S …(ii) [exterior angle of a triangle is greater than each of the opposite interior angles]

From (i) and (ii)

∠R > ∠S

PS > PR [side opposite to greater angle is longer]

⇒ PS > PQ [∵ PQ = PR]

Question 8.

In figure 3.51, in Î”ABC, seg AD and seg BE are altitudes and AE = BD.

Prove that seg AD ≅ seg BE

Given: AD and BE are altitudes

AE = BD

Proof: AD and BE are altitudes

∠ADB = ∠BEA = 90° [Given]

BD = AE [Given]

∠ADB = ∠BEA = 90° [Given]

AB = BA [Common side of both the triangles]

∴ By RHS congruency

So, AD ≅ BE [corresponding sides of congruent triangles]

###### Practice Set 3.5
Question 1.

If Î”XYZ ∼ Î”LMN write the corresponding angles of the two triangles and also write the ratios of corresponding sides.

Given, Î”XYZ ∼ Î”LMN

Corresponding angles of the two triangles are

∠X = ∠L

∠Y = ∠M

∠Z = ∠N

Ratios of corresponding sides.

Question 2.

In In Î”XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If Î”XYZ ∼ Î”PQR and PQ = 8 cm then find the lengths of remaining sides of Î”PQR.

Given,

In Î”XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm

Î”PQR, PQ = 8 cm

Î”XYZ ∼ Î”PQR

So, Ratios of corresponding sides.

⇒

⇒

⇒ QR = 6 × 2 cm and PR = 5 × 2 cm

⇒ QR = 12 cm and PR = 10 cm

Question 3.

Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same signs. Show the lengths of corresponding sides by numbers in proportion.

Î”ABC ∼ Î”XYZ

Corresponding Angles

∠A = ∠X

∠B = ∠Y

∠C = ∠Z

Corresponding Sides in proportion

###### Problem Set 3
Question 1.

Choose the correct alternative answer for the following questions.

If two sides of a triangle are 5 cm and 1.5 cm, the length of its third side cannot be ………..
A. 3.7 cm

B. 4.1 cm

C. 3.8 cm

D. 3.4 cm

The difference between two sides is less than third side
5 – 1.5 = 3.5

So, the third side cannot be 3.4 cm

Question 2.

Choose the correct alternative answer for the following questions.

In Î”PQR, If ∠R > ∠Q then ………….
A. QR>PR

B. PQ>PR

C. PQ<PR

D. QR<PR

∠R > ∠Q

∴ PQ > PR

Question 3.

Choose the correct alternative answer for the following questions.

In Î”TPQ, ∠T = 650, ∠P = 950 which of the following is a true statement?
A. PQ<TP

B. PQ<TQ

C. TQ<TP<PQ

D. PQ<TP<TQ

Sum of interior angles of a triangle = 180°

∠T + ∠P + ∠Q = 180°

⇒ 65° + 95° + ∠Q = 180°

⇒ ∠Q = 180° - 160° = 20°

Since, side opposite to greater angle is greater

∴ TP < PQ < TQ

Question 4.

Î”ABC is isosceles in which AB = AC. seg BD and seg CE are medians. Show that BD = CE.

Given: Î”ABC is an isosceles triangle.

BD and CE are medians.

AB = AC

1/2 AB = 1/2 AC

Since, 1/2 AB = BE = AE and 1/2 AC = AD = CD

So, BE = CD ………….(1)

Also, ∠ABC = ∠ACB

⇒ ∠EBC = ∠DCB ……….(2)

In Î”EBC and Î”DCB

BE = CD [from (1)]

∠EBC = ∠DCB [from (2)]

BC = CB [common side]

∴ By SAS congruency

Î”EBC ≅ Î”DCB

So,

CE = BD …………..corresponding sides of congruent triangles.

∴ BD = CE

Question 5.

In Î”PQR, If PQ>PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ>SR.

Given:

SQ and SR are bisectors of ∠Q and ∠R which meet at S

PQ > PR

To Prove: SQ > SR

Proof:

PQ > PR

∠PRQ > ∠PQR [angle opposite to longer side is larger] …………(1)

SQ and SR are bisectors of ∠Q and ∠R

∴ ∠SQR = 1/2 ∠PQR and ∠SRQ = 1/2 ∠PRQ

Dividing (1) by 1/2 we get

1/2 ∠PRQ > 1/2 ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SR [sides opposite to greater angle is longer]

Question 6.

In figure 3.59, point D and E are on side BC of Î”ABD, such that BD = CE and AD = AE. Show that Î”ABD ≅ Î”ACE.

Given: BD = CE

To Prove: Î”ABD ≅ Î”ACE

Proof:

⇒ ∠ADE = ∠AED [angles opposite to equal sides are equal] ……(1)

Subtracting 180° from (1)

⇒ 180° - ∠ADE = 180° - ∠AED

In Î”ABD and Î”ACE

BD = CE [Given]

∴ By SAS congruency test

Î”ABD ≅ Î”ACE

Question 7.

In figure 3.60, point S is any point on side QR of Î”PQR.

Prove that: PQ + QR + RP > 2PS

Given: S is any point on side QR of Î”PQR.

To Prove: PQ + QR + RP > 2PS

Proof:

We know, sum of two sides of triangle is greater than the third side

∴ In Î”PQS

PQ + QS > PS …………(1)

In Î” PSR

PR + SR > PS ……..(2)

PQ + QS + PR + SR > PS + PS

⇒ PQ + QS + SR + PR > 2PS

⇒ PQ + QR + PR > 2PS [QR = QS + SR]

Hence, proved.

Question 8.

In figure 3.61, bisector of ∠BAC intersects side BC at point D.

Prove that AB > BD

Given: AD is bisector of ∠BAC

To Prove: AB > BD

Proof: AD is bisector of ∠BAC

∠ADB > ∠DAC ..(2) [exterior angle of a triangle is greater than each

of the opposite interior angles]

Substituting ∠DAC = ∠BAD in (2)

⇒ AB > BD [side opposite to larger angle is larger]

Question 9.

In figure 3.62, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR

Given: PT is angle bisector of ∠QPR

⇒ ∠QPT = ∠RPT

A line through R parallel to PT intersects ray QP at S

RS || PT

To Prove: PS = PR

Proof:

PT is angle bisector of ∠QPR

⇒ ∠QPT = ∠RPT

∠QPR = ∠QPT + ∠RPT

∠QPR = 2∠RPT (1)

RS || PT, PR is the transversal

So, ∠RPT = ∠PRS [Alternate interior angles] (2)

For Î”PRS ∠RPQ is the remote exterior angle.

∠PSR + ∠PRS = ∠QPR

Substituting (1) and (2) in the above equation

∠RPT + ∠PSR = 2∠RPT

⇒ ∠PSR = ∠RPT (3)

From (2) and (3)

∠PRS = ∠PSR

⇒ PS = PR [Sides opposite to equal angles are equal]

Question 10.

In figure 3.63, seg AD ⊥ seg BC. seg AE is the bisector of ∠CAB and C - E - D. Prove that ∠DAE = 1/2 (∠B - ∠C)

Given: AE is bisector of ∠CAB.

To Prove: ∠DAE = 1/2 (∠B - ∠C)

Proof:

We know that ∠BAE = 1/2 ∠A (1)

∠BAD = 90° - ∠B ……………..(2)

On putting equations (1) and (2)

= 1/2 ∠A - (90° - ∠B)

= 1/2 ∠A – 90° + ∠B

= 1/2 ∠A - 1/2 (∠C + ∠A + ∠B) + ∠B

= 1/2 ∠A - 1/2 ∠A - 1/2 ∠B – 1/2 ∠C + ∠B

= 1/2 ∠B – 1/2 ∠C

∴ ∠DAE = 1/2 (∠B - ∠C)