Class 9 - Maths - Triangles

**Exercise 7.1**

**Question 1:**

In quadrilateral ACBD,AC = AD and AB bisects ∠ A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD.

What can you say about BC and BD?

Answer:

In Δ ABC and Δ ABD, we have

AC = AD [Given]

∠CAB = ∠DAB [Since AB bisects ∠A]

AB = AB [Common]

So, Δ ABC ≅ Δ ABD [By SAS congruence]

Therefore, BC = BD [By CPCT]

**Question 2:**

ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17).

Prove that

(i) Δ ABD ≅ Δ BAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC.

Answer:

(i) In quadrilateral ABCD, we have AD = BC and ∠DAB = ∠CBA.

In ΔABD and ΔBAC,

AD = BC [Given]

AB = BA [Common]

∠DAB = ∠CBA [Given]

Using SAS criteria, we have

ΔABD ≌ ΔBAC

(ii) Since ΔABD ≌ ΔBAC

So, their corresponding parts are equal.

=> BD = AC

(ii) Since ΔABD ≌ ΔBAC

So, their corresponding parts are equal.

=> ∠ABD = ∠BAC.

**Question 3:**

AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Answer:

We have ∠ABC = 90^{0} and ∠BAD = 90^{0}

Also AB and CD intersect at O.

So, vertically opposite angles are equal.

Now, in ΔOBC and ΔOAD, we have

∠ABC = ∠BAD [each = 90^{0}]

BC = AD [Given]

∠BOC = ∠AOD [vertically opposite angles]

Using ASA criteria, we have

ΔOBC ≌ ΔOAD

=> OB = OA [By CPCT]

i.e. O is the mid-point of AB

Thus, CD bisects AB.

**Question 4:**

l and m are two parallel lines intersected by another pair of parallel lines p and q

(see Fig. 7.19). Show that Δ ABC ≅ Δ CDA.

Answer:

Since l || m and AC is a transversal.

So, ∠BAC = ∠DCA [Alternate interior angles]

Also p || q and AC is a transversal,

So, ∠BCA = ∠DAC [Alternate interior angles]

Now, in ΔABC and ΔCDA,

∠BAC = ∠DCA [Proved]

∠BCA = ∠DAC [Proved]

CA = AC [Common]

Using ASA criteria, we have

ΔABC ≌ ΔCDA

**Question 5:**

line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Fig. 7.20).

Show that:

(i) Δ APB ≅ Δ AQB

(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Answer:

We have, l as the bisector of QAP.

So, ∠QAB = ∠PAB

∠Q = ∠P [each = 90^{0}]

=> Third ∠ABQ = Third ∠ABP

(i) Now, in ΔAPB and ΔAQB, we have

AB = AB [Common]

∠ABP = ∠ABQ [Proved]

∠PAB = ∠QAB [Proved]

Using SAS criteria, we have

ΔAPB ≌ ΔAQB

(ii) Since ΔAPB ≌ ΔAQB

So, their corresponding angles are equal.

=> BP = BQ

i.e. Perpendicular distance of B from AP = Perpendicular distance of B from AQ

Thus, the point B is equidistant from the arms of ∠A.

**Question 6:**

In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.

Answer:

We have ∠BAD = ∠EAC

Adding ∠DAC on both sides, we have

∠BAD + ∠DAC = ∠EAC + ∠DAC

=> ∠BAC = ∠DAE

Now, in ΔABC and ΔADE, we have

∠BAC = ∠DAE [Proved]

AB = AD [Given]

AC = AE [Given]

So, ΔABC ≌ ΔADE [Using SAS criteria]

Since ΔABC ≌ ΔADE, therefore, their corresponding parts are equal.

=> BC = DE.

**Question 7:**

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that

∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB

(see Fig. 7.22). Show that

(i) Δ DAP ≅ Δ EBP

(ii) AD = BE

Answer:

We have, P is the mid-point of AB.

AP = BP

∠EPA = ∠DPB [Given]

Adding ∠EPD on both sides, we get

∠EPA + ∠EPD = ∠DPB + ∠EPD

∠APD = ∠BPE

(i) Now, in ΔDAP ≌ ΔEBP, we have

AP = BP [Proved]

∠PAD = ∠ PBE [It is given that ∠BAD = ∠ABE]

∠DPA = ∠EPB [Proved]

Using ASA criteria, we have

ΔDAP ≌ ΔEBP

(ii) Since ΔDAP ≌ ΔEBP

So, their corresponding parts are equal.

=> AD = BE.

Q**uestion 8:**

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that

DM = CM. Point D is joined to point B (see Fig. 7.23).

Show that:

(i) Δ AMC ≅ Δ BMD

(ii) ∠ DBC is a right angle.

(iii) Δ DBC ≅ Δ ACB

(iv) CM = AB/2

Answer:

Since M is the mid-point of AB.

So, BM = AM [Given]

(i) In ΔAMC and ΔBMD, we have

CM = DM [Given]

AM = BM [Proved]

∠AMC = ∠BMD [Vertically opposite angles]

So, ΔAMC ≌ ΔBMD [SAS criteria]

(ii)Since ΔAMC ≌ ΔBMD

So, their corresponding parts are equal.

=> ∠MAC = ∠MBD

But they form a pair of alternate interior angles.

So, AC || DB

Now, BC is a transversal which intersecting parallel lines AC and DB,

So, ∠BCA + ∠DBC = 180^{0}

But ∠BCA = 90^{0} [Since ΔABC is right angled at C]

=> 90^{0} + ∠DBC = 180^{0}

=> ∠DBC = 180^{0} – 90^{0}

=> ∠DBC = 90^{0}

(iii) Again, ΔAMC ≌ ΔBMD [Proved]

So, AC = BD [by CPCT]

Now, in ΔDBC and ΔACB, we have

∠DBC = ∠ACB [Each = 90^{0}]

BD = CA [Proved]

BC = CB [Common]

Using SAS criteria, we have

ΔDBC ≌ ΔACB

(iv) Since ΔDBC ≌ ΔACB

So, their corresponding parts are equal.

=> DC = AB

But DM = CM [Given]

So, CM = DC/2 = AB/2

=> CM = AB/2

**Exercise 7.2**

**Question 1:**

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠ A

Answer:

(i) In ΔABC, we have

AB = AC [Given]

∠C = ∠B [Angle opposite to equal sides are equal]

∠OCB = ∠OBC

=> OB = OC [Sides opposite to equal angles are equal

(ii) In ΔABO and ΔACO, we have

AB = AC [Given]

OB = OC [Proved]

Using SAS criteria,

ΔABO ≌ ΔACO

=> ∠OAB = ∠OAC [By CPCT]

=> AO bisects ∠A.

**Question 2:**

In Δ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Δ ABC is an isosceles triangle in which AB = AC.

Answer:

Since AD is bisector of BC.

So, BD = CD

Now, in ΔABD and ΔACD, we have:

AD = AD [Common]

∠ADB = ∠ADC = 90^{0} [Since AD ⊥ BC]

BD = CD [Proved]

So, ΔABD ≌ ΔACD [SAS criteria]

Hence, their corresponding parts are equal.

=> AB = AC

Thus, ΔABC is an isosceles triangle.

**Question 3:**

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31).

Show that these altitudes are equal.

Answer:

ΔABC is an isosceles triangle. So, AB = AC

=> ∠ACB = ∠ABC [Since angles opposite to equal sides are equal]

Now, in ΔBEC and ΔCFB, we have

∠EBC = ∠FCB [Proved]

BC = CB [Common]

and ∠BEC = ∠CFB [Each = 90^{0}]

So, ΔBEC ≌ ΔCFB [Using ASA criteria]

=> Their corresponding parts are equal. i.e. BE = CF

**Question 4: **

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32).

Show that

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer:

(i) In ΔABE and ΔACF, we have

∠AEB = ∠AFC [each = 90^{0} since BE ⊥AC and CF ⊥ AB]

∠A = ∠A [Common]

BE = CF [Given]

So, ΔABE ≌ ΔACF [Using AAS criterion]

(ii) Since, ABE ≌ ΔACF

So, their corresponding parts are equal.

=> AB = AC

**Question 5: **

ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ ABD = ∠ ACD.

Answer:

In ΔABC, we have

AB = AC [Since ΔABC is an isosceles triangle]

But angles opposite to equal sides are equal.

So, ∠ABC = ∠ACB ……....(1)

Again, in ΔBDC, we have

BD = CD [Since ΔBDC is an isosceles triangle.]

So, ∠CBD = ∠BCD …....(2) [Angles opposite to equal sides are equal]

Adding equation (1) and (2), we have

∠ABC + ∠CBD = ∠ACB + ∠BCD

=> ∠ABD = ∠ACD

**Question 6:**

ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB

(see Fig. 7.34). Show that ∠ BCD is a right angle.

Answer:

In ΔABC,

AB = AC [Given] …....(1)

AB = AD [Given] …....(2)

From equation (1) and (2), we have

AC = AD

Now, in ΔABC, we have

∠B + ∠ACB + ∠BAC = 180^{0}

=> 2∠ACB + ∠BAC = 180^{0} ….....(3) [Since ∠B = ∠ACB (Angles opposite to equal sides)]

In ΔACD,

∠D + ∠ACD + ∠CAD = 180^{0}

=> 2∠ACD + ∠CAD = 180^{0} …….(4) [Since ∠D = ∠ACD (angles opposite to equal sides)]

Adding equations (3) and (4), we have

2∠ACB + ∠BAC + 2∠ACD + ∠CAD = 180^{0} + 180^{0}

=> 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360^{0}

=> 2[∠BCD] + [180^{0}] = 360^{0} [∠BAC and ∠CAD form a linear pair]

=> 2∠BCD = 360^{0} – 180^{0} = 180^{0}

=> ∠BCD = 180^{0}/2 = 90^{0}

Thus, ∠BCD = 90^{0}

**Question 7:**

ABC is a right angled triangle in which ∠ A = 90^{0 }and AB = AC. Find ∠ B and ∠ C.

Answer:

In ΔABC, we have

AB = AC [Given]

So, their opposite angles are equal.

=> ∠ACB = ∠ABC

Now, ∠A + ∠B + ∠C = 180^{0}

=> 90^{0} + ∠B + ∠C = 180^{0} [Since ∠A = 90^{0} (Given)]

=> ∠B + ∠C = 180^{0} - 90^{0}

=> ∠B + ∠C = 90^{0}

But ∠ABC = ∠ACB, i.e. ∠B = ∠C

Thus, ∠B = 45^{0} and ∠C = 45^{0}

**Question 8:**

Show that the angles of an equilateral triangle are 60^{0} each.

Answer:

In ΔABC, we have

AB = BC = CA [Since ABC is an equilateral triangle]

=> AB = BC

=> A = ∠C ……...(1) [Since Angle opposite to equal sides are equal.]

Similarly, AC = BC

=> ∠A = ∠B …....(2)

From equation (1) and (2), we have

∠A = ∠B = ∠C

Let ∠A = ∠B = ∠C = x

Since, ∠A + ∠B + ∠C = 180^{0}

=> x + x + x = 180^{0}

=> 3x = 180^{0}

=> x = 180^{0}/3

=> x = 60^{0}

Thus, the angles of an equilateral triangle are 60^{0} each.

**Exercise 7.3**

**Question 1:**

Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the

same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that

(i) Δ ABD ≅ Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC.

Answer:

(i) In ΔABD and ΔACD, we have

AB = AC

AD = AD

BD = CD

ΔABD ≌ ΔACD

(ii) In ΔABP and ΔACP, we have

AB = AC [Given]

So, AB = AC=> ∠B = ∠C [Since angle opposite to equal sides are equal]

AP = AP [Common]

So, ΔABP ≌ ACP [SAS Criteria]

(iii) Since, ΔABP ≌ ΔACP

So, their corresponding parts are congruent.

=> ∠BAP = ∠CAP

Hence, AP is the bisector of ∠A. …...(1)

Again, in ΔBDP and ΔCDP, we have

BD = CD [Given]

∠DBP = ∠CDP [Angles opposite to equal sides]

DP = DP [Common]

=> ∠BDP ≌ ∠CDP

So, ∠BDP ≌ ∠CDP [By CPCT]

Hence, DP (or AP) is the bisector of ∠D. …...(2)

From equation (1) and (2), AP is the bisector of ∠A as well as ∠D.

(iv) Since ΔABP ≌ ΔACP

So, their corresponding parts are equal.

=> ∠APB = ∠APC

But ∠APB + ∠APC = 180^{0} [Linear pair]

So, ∠APB = ∠APC = 90^{0}

=> AP ⊥ BC

=> AP is the perpendicular bisector of BC

**Question 2:**

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ∠ A.

Answer:

(i) In ΔABD and ΔACD, we have

AB = AC [Given]

∠B = ∠C [Angles opposite to equal sides]

AD = AD [Common]

So, ΔABD ≌ ΔACD

Hence, their corresponding parts are equal.

=> BD = CD

=> D is the mid-point of BC or AD bisects BC.

(ii) Since, ΔABD ≌ ΔACD,

So, their corresponding parts are congruent.

=> ∠BAD = ∠CAD

=> AD bisects ∠A.

**Question 3:**

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40).

Show that:

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Answer:

In ΔABC, AM is a median

So, BM = BC/2 [Given] …....(1)

In ΔPQR, PN is a median.

So, QN = QR/2

=> BC = QR …………2

BM =QN [From equation (1) and (2)]

(i) In ΔABM and ΔPQN, we have

AB = PQ [Given]

AM = PN [Given]

BM = QN [Proved]

So, ΔABM ≌ ΔPQN [SSS criteria]

(ii)Since ΔABM ≌ ΔPQN

So, their corresponding parts are congruent.

=> ∠B = ∠Q

Now, in ΔABC and ΔPQR, we have

∠B = ∠Q [Proved]

AB = PQ [Given]

BC = QR [Given]

So, ΔABC ≌ ΔPQR

**Question 4:**

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:

Given, BE ⊥ AC

So, ΔBEC is a right triangle such that ∠BEC = 90^{0}

Similarly, ∠CFB = 90^{0}

Now, in right ΔBEC and right ΔCFB, we have

BE = CF [Given]

BC = CB [Common]

Using RHS criteria, ΔBEC ≌ ΔCFB

So, their corresponding parts are equal.

=> ∠BCE = ∠CBF or ∠BCA = ∠CBA

Now, in ΔABC, ∠BCA = ∠CBA

So, their opposite sides are equal.

=> AB = AC

Hence, ΔABC is an isosceles triangle.

**Question 5:**

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.

Answer:

We have AP ⊥ BC [Given]

So, ∠APB = 90^{0} and APC = 90^{0}

In ΔABP and ΔACP, we have

∠APB = ∠APC [each = 90^{0}]

AB = AC [Given]

AP = AP [Common]

Using RHS criteria,

ΔABP ≌ ΔACP

So, their corresponding parts are congruent.

=> ∠B = ∠C

**Exercise 7.4**

**Question 1:**

Show that in a right angled triangle, the hypotenuse is the longest side.

Answer:

Let us consider ΔABC such that ∠B = 90^{0}

So, ∠A + ∠B + ∠C = 180^{0}

=> [∠A + ∠C] + ∠B = 180^{0}

=> ∠A + ∠C = ∠B

=> ∠B > ∠A and ∠B > ∠C

=> Side opposite to ∠B is longer than the side opposite to ∠A.

i.e. AC > BC …....(1)

Similarly, AC > AB …....(2)

From equation (1) and (2), we get,

AC is the longest side.

But AC is the hypotenuse of the triangle.

Thus, hypotenuse is the longest side.

**Question 2: **

In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively.

Also, ∠ PBC < ∠ QCB. Show that AC > AB.

Answer:

∠ABC + ∠PBC = 180^{0} [Linear pair]

∠ACB + ∠QCB = 180^{0} [Linear pair]

So, ∠ABC + ∠PBC = ∠ACB + ∠QCB

But ∠PBC < ∠QCB [Given]

So, ∠ABC > ∠ACB

=> [The side opposite to ∠ABC] > [The side opposite to ∠ACB]

=> AC > AB

**Question 3: **

In Fig. 7.49, ∠ B < ∠ A and ∠ C < ∠ D.Show that AD < BC.

Answer:

Given ∠B < ∠A

=> ∠A > ∠B

So, OB > OA [Since side opposite to greater angle is longer]........(1)

Similarly,

OC > OD............(2)

From equation (1) and (2), we have

[OB + OC] > [OA + OD]

=> BC > AD

**Question 4: **

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD

(see Fig. 7.50). Show that ∠ A > ∠ C and ∠ B > ∠ D.

Answer:

Let us join AC.

Now, in ΔABC,

AB < BC [Since AB is the smallest side of quadrilateral ABCD]

=> BC > AB

=> [Angle opposite to BC] < [Angle opposite to AB]

=> ∠BAC > ∠BCA ...........(1)

Again, in ΔACD,

CD > AD [Since CD is the longest side of the quadrilateral ABCD]

=> [Angle opposite to CD] > [Angle opposite to AD]

=> ∠CAD > ∠ACD ……....(2)

Adding equation (1) and (2), we get

[∠BAC + CAD] > [∠BCA + ∠ACD]

=> ∠A > ∠C

Similarly, by joining BD, we have

∠B > ∠D

**Question 5: **

In Fig 7.51, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ

Answer:

In ΔPAR, PS bisects ∠QPR [Given]

So, ∠QPS = ∠RPS

Again PR > PQ [Given]

=> [Angle opposite to PR] > [Angle opposite to PQ]

=> ∠PQS > ∠PRS

=> [∠PQS + ∠QPS] > [∠PRS + ∠RPS] …....(1) [Since ∠QPS = ∠RPS]

So, Exterior ∠PSR = [∠PQS + ∠QPS]

[An exterior angle is equal to the sum of interior opposite angles]

And Exterior ∠PSQ = [∠PRS + ∠RPS]

Now, from (1), we have

∠PSR > ∠PSQ

**Question 6:**

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer:

Let we have a line l and O is a point not on line l such that OP Ʇ l.

We have to prove that OP < OQ, OP < OR and OP < OS.

In ΔOPQ,

∠P = 90^{0}

So, ∠Q is an acute angle i.e. ∠Q < 90^{0}

So, ∠Q < ∠P

Hence, OP < OQ [Side opposite to greater angle is longer]

Similarly, we can prove that OP is shorter than OR, OS.

**Exercise 7.5 (Optional)**

**Question 1:**

ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of Δ ABC.

Answer:

Given, ABC is a triangle. Now draw perpendicular bisectors of sides AB, BC and CA which meets at point O.

Hence O is the required point.

**Question 2:**

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer:

Let us consider a ΔABC.

Draw ‘l’ the bisector of ∠B.

Draw ‘m’ the bisector of ∠C.

Let the two bisectors l and m meet at O.

Thus, ‘O’ is the required point which is equidistant from the sides of ΔABC.

**Question 3:**

In a huge park, people are concentrated at three points (see Fig. 7.52):

A: where there are different slides and swings for children,

B: near which a man-made lake is situated,

C: which is near to a large parking and exit.

Where should an icecream parlour be set up so that maximum number of persons can approach it?

(Hint : The parlour should be equidistant from A, B and C)

Answer:

Let us join A and B, and draw ‘l’ the perpendicular bisector of AB.

Now, join B and C, and draw ‘m’ the perpendicular bisector of BC.

Let the perpendicular bisectors ‘l’ and ‘m’ meet at ‘O’.

The point ‘O’ is the required point where the ice cream parlour be set up.

Note: If we join ‘A’ and ‘C’, and draw the perpendicular bisectors, then it will also meet (or

pass through) the point O.

**Question 4:**

Complete the hexagonal and star shaped Rangolies [see Fig. 7.53 (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can.

Count the number of triangles in each case. Which has more triangles?

Answer:

It is an activity. We get the 150 equilateral triangles in the figure (i) and 300 equilateral

triangles in the figure (ii).

Hence, the figure (ii) has more triangles.

.