Important Questions for CBSE Class 9 Maths Chapter 7

CBSE Class 9 Maths Chapter-7 Important Questions - Free PDF Download

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1 Marks Quetions

1. In fig, if AD =BC and BAD =ABC, then ACB is equal to

(A) ABD

(B) BAD

(C) BAC

(D) BDA

Ans. (D) BDA

2. IN fig, if ABCD is a quadrilateral in which AD= CB, AB=CD, and D=B, then CAB is equal to

(A) ACD

(B) CAD

(C) ACD

(D) BAD

Ans. (C) ACD

3. If O is the mid – point of AB and BQO =APO, then OAP is equal to

(A) QPA

(B) OQB

(C) QBO

(D) BOQ

Ans. (C) QBO

4. IF AB BC and A =c, then the true statement is

(A) ABAC

(B) AB=BC

(C) AB=AD

(D) AB=AC

Ans. (B) AB=BC

5. If ABC is an isosceles triangle and B =, find x.

(a)

(b)

(c)

(d) none of these

Ans. (c)

6. If AB=AC and ACD=, find A

(a)

(b)

(c)

(d) none of these

Ans. (b)

7. What is the sum of the angles of a quadrilateral:

(a)

(b)

(c)

(d)

Ans. (b)

8. The sum of the angles of a triangle will be:

(a)

(b)

(c)

(d)

Ans. (c)

9. An angle is more than its complement. Find its measure.

(A) 42

(B) 32

(C) 52

(D) 62

Ans. (C) 52

10. An angle is 4 time its complement. Find measure.

(A) 62

(B) 72

(C) 52

(D) 42

Ans. (B) 72

11. Find the measure of angles which is equal to its supplementary.

(A)

(B)

(C)

(D)

Ans. (D)

12. Which of the following pairs of angle are supplementary?

(A)

(B)

(C)

(D) None of these.

Ans. (B)

13. Find the measure of each exterior angle of an equilateral triangle.

(A)

(B)

(C)

(D)

Ans. (C)

14. In an isosceles ABC, if AB=AC and, Find B.

(A)

(B)

(C)

(D)

Ans. (A)

15. In a ABC, if B=C=, Which is the longest side.

(A) BC

(B) AC

(C) CA

(D) None of these.

Ans. (A) BC

16. In a ABC, if AB=AC and B=, Find A.

(A)

(B)

(C)

(D)

Ans. (A)

17. Determine the shortest sides of the triangles.

(a) AC

(b) BC

(c) CA

(d) none of these

Ans. (b) BC

18. In an ABC, if = and =, determine the longest sides of the triangle.

(a) AC

(b) CA

(c) BC

(d) none of these

Ans. (a) AC

19. The sum of two angles of a triangle is equal to its third angle. Find the third angles.

(a)

(b)

(c)

(d)

Ans. (a)

20. Two angles of triangles are respectively. Find third angles.

(a)

(b)

(c)

(d)

Ans. (d)

21. ABC is an isosceles triangle with AB=AC and =, find.

Ans.

[angle opposite to equal sides are equal]

But,

And,

22. 1. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB (See figure)

Ans. In BOC and AOD,

OBC = OAD = [Given]

BOC = AOD [Vertically Opposite angles]

BC = AD [Given]

BOC AOD [By ASA congruency]

OB = OA and OC = OD [By C.P.C.T.]

2 Marks Quetions

1. In quadrilateral ABCD (See figure). AC = AD and AB bisects A. Show that ABC ABD. What can you say about BC and BD?

Ans. Given: In quadrilateral ABCD, AC = AD and AB bisects A.
To prove: ABC ABD
Proof: In ABC and ABD,
AC = AD [Given]
BAC = BAD [ AB bisects A]
AB = AB [Common]
ABC ABD [By SAS congruency]
Thus BC = BD [By C.P.C.T.]

2. ABCD is a quadrilateral in which AD = BC and DAB = CBA. (See figure). Prove that:

(i) ABD BAC

(ii) BD = AC

(iii) ABD = BAC

Ans. (i) In ABC and ABD,

BC = AD [Given]

DAB = CBA [Given]

AB = AB [Common]

ABC ABD [By SAS congruency]

Thus AC = BD [By C.P.C.T.]

(ii) Since ABC ABD

AC = BD [By C.P.C.T.]

(iii) Since ABC ABD

ABD = BAC [By C.P.C.T.]

3. and are two parallel lines intersected by another pair of parallel lines and (See figure). Show that ABC CDA.

Ans. AC being a transversal. [Given]

Therefore DAC = ACB [Alternate angles]

Now [Given]

And AC being a transversal. [Given]

Therefore BAC = ACD [Alternate angles]

Now In ABC and ADC,

ACB = DAC [Proved above]

BAC = ACD [Proved above]

AC = AC [Common]

ABC CDA [By ASA congruency]

4. Line is the bisector of the angle A and B is any point on BP and BQ are perpendiculars from B to the arms of A. Show that:

(i) APB AQB

(ii) BP = BQ or P is equidistant from the arms of A (See figure).

Ans. Given: Line bisects A.

BAP = BAQ

(i) In ABP and ABQ,

BAP = BAQ [Given]

BPA = BQA = [Given]

AB = AB [Common]

APB AQB [By ASA congruency]

(ii) Since APB AQB

BP = BQ [By C.P.C.T.]

B is equidistant from the arms of A.

5. In figure, AC = AB, AB = AD and BAD = EAC. Show that BC = DE.

Ans. Given that BAD = EAC

Adding DAC on both sides, we get

BAD + DAC = EAC + DAC

BAC = EAD ……….(i)

Now in ABC and AED,

AB = AD [Given]

AC = AE [Given]

BAC = DAE [From eq. (i)]

ABC ADE [By SAS congruency]

BC = DE [By C.P.C.T.]

6. AB is a line segment and P is the mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that:

(i) DAFFBPE D

(ii) AD = BE (See figure)

Ans. Given that EPA = DPB
Adding EPD on both sides, we get
EPA + EPD = DPB + EPD
APD = BPE ……….(i)
Now in APD and BPE,

PAD = PBE [BAD = ABE (given), PAD = PBE]
AP = PB [P is the mid-point of AB]
APD = BPE [From eq. (i)]
DPAEBP [By ASA congruency]
AD = BE [ ByC.P.C.T.]

7. 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.

Ans. (i) ABC is an isosceles triangle in which AB = AC.

C = B [Angles opposite to equal sides]

OCA + OCB = OBA + OBC

OB bisects B and OC bisects C

OBA = OBC and OCA = OCB

OCB + OCB = OBC + OBC

2OCB = 2OBC

OCB = OBC

Now in OBC,

OCB = OBC [Prove above]

OB = OC [Sides opposite to equal sides]

(ii) In AOB and AOC,

AB = AC [Given]

OBA = OCA [Given]

And B = C

B = C

OBA = OCA

OB = OC [Prove above]

AOBAOC [By SAS congruency]

OAB = OAC [By C.P.C.T.]

Hence AO bisects A.

8. In ABC, AD is the perpendicular bisector of BC (See figure). Show that ABC is an isosceles triangle in which AB = AC.

Ans. In AOB and AOC,

BD = CD [AD bisects BC]

ADB = ADC = [AD BC]

AD = AD [Common]

ABD ACD [By SAS congruency]

AB = AC [By C.P.C.T.]

Therefore, ABC is an isosceles triangle.

9. ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (See figure). Show that these altitudes are equal.

Ans. In ABE and ACF,

A= A [Common]

AEB = AFC = [Given]

AB = AC [Given]

ABE ACF [By ASA congruency]

BE = CF [By C.P.C.T.]

Altitudes are equal.

10. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (See figure). Show that:

(i) ABE ACF

(ii) AB = AC or ABC is an isosceles triangle.

Ans. (i) In ABE and ACF,

A= A [Common]

AEB = AFC = [Given]

BE = CF [Given]

ABE ACF [By ASA congruency]

(ii) Since ABE ACF

BE = CF [By C.P.C.T.]

ABC is an isosceles triangle.

11. ABC and DBC are two isosceles triangles on the same base BC (See figure). Show that ABD = ACD.

Ans. In isosceles triangle ABC,

AB = AC [Given]

ACB = ABC …….(i) [Angles opposite to equal sides]

Also in Isosceles triangle BCD.

BD = DC

BCD = CBD ……….(ii) [Angles opposite to equal sides]

Adding eq. (i) and (ii),

ACB + BCD = ABC + CBD

ACD = ABD

Or ABD = ACD

12. ABC is a right angled triangle in which A = and AB = AC. Find B and C.

Ans. ABC is a right triangle in which,

A = And AB = AC

In ABC,

AB = AC C = B ……….(i)

We know that, in ABC, A + B + C = [Angle sum property]

B + B = [A = (given) and B = C (from eq. (i)]

2B =

B =

Also C = [B = C]

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

(i) AD bisects BC.

(ii) AD bisects A.

Ans. In ABD and ACD,

AB = AC [Given]

ADB = ADC = [AD BC]

AD = AD [Common]

ABD ACD [RHS rule of congruency]

BD = DC [By C.P.C.T.]

AD bisects BC

Also BAD = CAD [By C.P.C.T.]

AD bisects A.

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

Ans. Given: Let ABC be a right angled triangle, right angled at B.

To prove: Hypotenuse AC is the longest side.

Proof: In right angled triangle ABC,

A + B + C =

A + + C = [B = ]

A + C =

And B =

B >C and B >A

Since the greater angle has a longer side opposite to it.

AC > AB and AC > AB

ThereforeB being the greatest angle has the longest opposite side AC, i.e. hypotenuse.

15. In figure, sides AB and AC of ABC are extended to points P and Q respectively. Also PBC<QCB. Show that AC > AB.

Ans. Given: In ABC, PBC<QCB

To prove: AC > AB

Proof: In ABC,

4 >2 [Given]

Now 1 + 2 = 3 + 4 = [Linear pair]

1 >3 [4 >2]

AC > AB [Side opposite to greater angle is longer]

16. In figure, B <A and C <D. Show that AD < BC.

Ans. In AOB,

B <A [Given]

OA < OB ……….(i) [Side opposite to greater angle is longer]

In COD,

C <D [Given]

OD < OC ……….(ii) [Side opposite to greater angle is longer]

Adding eq. (i) and (ii),

OA + OD < OB + OC

AD < BC

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

Ans. Let ABC be a triangle.

Draw bisectors of B and C.

Let these angle bisectors intersect each other at point I.

Draw IK BC

Also draw IJ AB and IL AC.

Join AI.

In BIK and BIJ,

IKB = IJB = [By construction]

IBK = IBJ

[ BI is the bisector of B (By construction)]

BI = BI [Common]

BIK BIJ [ASA criteria of congruency]

IK = IJ [By C.P.C.T.] ……….(i)

Similarly, CIKCIL

IK = IL [By C.P.C.T.] ……….(ii)

From eq (i) and (ii),

IK = IJ = IL

Hence, I is the point of intersection of angle bisectors of any two angles of ABC equidistant from its sides.

18. In quadrilateral ACBD, AB=AD and AC bisects A. show ABC ACD?

Ans. IN ABC and ACD,

AD=AB……… (Given)

BAC= CAD……. (AC bisectsA)

And AC= AC ………….. (Common)

ABC ACD ……….. (SAS axiom)

19. If DA and CB are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Ans. In AOD and BOC,

AD=BC ……….. (Given)

And (vert opp. Angles)

(AAS rule)

(CPCT)

Hence CD bisects AB.

20. l and m, two parallel lines, are intersected by Another pair of parallel lines p and C. show that ABC CDA.

Ans. cuts them – (Given)

(alternate angles)

(Given)

(Alternate angles)

AC=CA (common)

(ASA rule)

21. In fig, the bisector AD of ABC is to the opposite side BC at D. show that ABC is isosceles?

Ans. In ABD and ACD

22. If AE=AD and BD=CE. Prove that AEB ADC

Ans. We have,
AE=AD and CE=BD
AE+CE=AD+BD
AC=AB(i)
Now, in AEB and ADC,
AE=AD [given]
EAB=DAC [common]
AB=AC [from (i)]
AEBADC [by SAS]

23. In quadrilateral ACBD, AC=AD and AB bisects A. show that ABCABD. What can you say about BC and BD?

Ans. In ABC and ABD,
AC=AD [given]
CAB=DAB[AB bisects A]
AB=AB [common]
ABCABD[SAS criterion]
BC=BD [CPCT]

24. In ABC, the median AD isto BC. Prove that ABC is an isosceles triangle.

Ans.
BD =CD [D is mid-point of BC]
AD=AD [Common]

[each ]
[By SAS]
[CPCT]
Hence, triangle ABC is an isosceles triangle.

25. Prove that ABC is isosceles if altitude AD bisects BAC.

Ans.

AD=AD [common]
[By AAs]
[CPCT]
Thus, is an isosceles triangle.

26. ABC is An isosceles triangle in which altitudes BE and CF are drawn to side AC and AB respectively. Show that these altitudes are equals.

Ans.

AB=AC [given]
[AAS rule]
[CPCT]

27. If AC= AE, AB=AD and = show that BC =DE.

Ans.
AB=AD [given]
AC=AE [given]
Also, [given]

[SAS criterior]
[CPCT]

28. Line is the bisector of an angle A and B is any point on line l. BP and BQ are from B to the arms of A show that :

(i) APB AQB

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

Ans.

[common]
AB=AB [Common]

[AAS rule]
[CPCT]

29. In the given figure, ABC is an isosceles triangle and B =, find x.

Ans.
AB=AC
[Angles opposite to equal sides are equal]
But

So,

30. If E>A and C>D. prove that AD>EC.

Ans.

[given]
[Side opposite to greater angle is larger] …..(i)
Similarly, in

Adding (i) and (ii)

31. If PQ= PR and S is any point on side PR. Prove that RS<QS.

Ans.
PQ=PR [given]
[angle opposite to equal side are equal]
Now,

[side opposite to smaller angle in ]

32. Prove that MN+NO +OP+PM>2MO.

Ans.

MN+NO>MO [Sum of any two side of is greater than third sides] …(i)

Similarly in

OP+PM>MO ….(ii)

Hence from (i) and (ii)

Or MN+NO+OP+PM>2MO

33. Prove that MN+NO+OP>PM.

Ans.

MN+NO>MO [Sum of any two side of is greater than third sides] …(i)
Similarly in
MO+OP>PM ….(ii)
Hence from (i) and (ii)
Or MN+NO+OP+MO>MO+PM
Or MN+NO+OP>PM

34. ABC is an equilateral triangle and =, find.

Ans.
AB=AC
[angles opposite to equal sides are equal]
But
So,

35. In the figure, AB = AC and.

Ans.
[angles opposite to equal sides are equal]
Also, [Linear pair]

and,

36. In the given figure, find

Ans.
[sum of three angles of a]

3 Marks Quetions

1. Prove that in a right triangle, hypotenuse is the longest (or largest) side.

Ans. Given a right angled triangle ABC in which

Now, since

Hence, the side opposite to is the hypotenuse and the longest side of the triangle.

2. Show that the angles of an equilateral triangle are 60^o each.

Ans. Let ABC be an equilateral triangle.

AB = BC = ACAB = BC

C = A……….(i)

Similarly, AB = AC

C = B……….(ii)

From eq. (i) and (ii),

A = B = C……….(iii)

Now in ABC

A + B + C = ……….(iv)

A + A + A = 3A =

A =

SinceA = B = C[From eq. (iii)]

A = B = C =

Hence each angle of equilateral triangle is

3. ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (See figure). If AD is extended to intersect BC at P, show that:

(i)ABD ACD

(ii) ABPACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

Ans. i)ABC is an isosceles triangle.

AB = AC

DBC is an isosceles triangle.

BD = CD

Now in ABD and ACD,

AB = AC[Given]

BD = CD[Given]

AD = AD[Common]

ABD ACD[By SSS congruency]

BAD = CAD[By C.P.C.T.]……….(i)

(ii)Now in ABP and ACP,

AB = AC[Given]

BAD = CAD[From eq. (i)]

AP = AP

ABPACP[By SAS congruency]

(iii)SinceABPACP[From part (ii)]

BAP = CAP[By C.P.C.T.]

AP bisects A.

SinceABD ACD[From part (i)]

ADB = ADC[By C.P.C.T.]……….(ii)

NowADB + BDP = [Linear pair]……….(iii)

AndADC + CDP = [Linear pair]……….(iv)

From eq. (iii) and (iv),

ADB + BDP = ADC + CDP

ADB + BDP = ADB + CDP[Using (ii)]

BDP = CDP

DP bisects DorAP bisects D.

(iv) SinceABPACP[From part (ii)]

BP = PC[By C.P.C.T.]……….(v)

AndAPB = APC[By C.P.C.T.]……….(vi)

NowAPB + APC = [Linear pair]

APB + APC = [Using eq. (vi)]

2APB =

APB =

AP BC……….(vii)

From eq. (v), we have BP PC and from (vii), we have proved AP B. So, collectively AP is perpendicular bisector of BC.

4. Two sides AB and BC and median AM of the triangle ABC are respectively equal to side PQ and QR and median PN of PQR (See figure). Show that:

(i)ABM PQN

(ii)ABC PQR

Ans. AM is the median of ABC.

BM = MC = BC……….(i)

PN is the median of PQR.

QN = NR = QR……….(ii)

Now BC = QR[Given] BC = QR

BM = QN……….(iii)

(i)Now in ABM and PQN,

AB = PQ[Given]

AM = PN[Given]

BM = QN[From eq. (iii)]

ABM PQN[By SSS congruency]

B = Q[By C.P.C.T.]……….(iv)

(ii)In ABC and PQR,

AB = PQ[Given]

B = Q[Prove above]

BC = QR[Given]

ABC PQR[By SAS congruency]

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

Ans. In BEC and CFB,

BEC = CFB[Each ]

BC = BC[Common]

BE = CF[Given]

BECCFB[RHS congruency]

EC = FB[By C.P.C.T.]…..(i)

Now In AEB and AFC

AEB = AFC [Each ]

A = A[Common]

BE = CF[Given]

AEBAFC[ASA congruency]

AE = AF[By C.P.C.T.]…………(ii)

Adding eq. (i) and (ii), we get,

EC + AE = FB + AFAB = AC

ABC is an isosceles triangle.

6. ABC is an isosceles triangles with AB = AC. Draw AP BC and show that B = C.

Ans. Given: ABC is an isosceles triangle in which AB = AC

To prove:B = C

Construction: Draw AP BC

Proof: In ABP and ACP

APB = APC = [By construction]

AB = AC[Given]

AP = AP[Common]

ABPACP[RHS congruency]

B = C[By C.P.C.T.]

7. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (See figure). Show that A >C and B >D.

Ans. Given: ABCD is a quadrilateral with AB as smallest and CD as longest side.

To prove:(i)A >C(ii) B >D

Construction: Join AC and BD.

Proof:(i)In ABC, AB is the smallest side.

4 <2 ……….(i)

[Angle opposite to smaller side is smaller]

In ADC, DC is the longest side.

3 <1 ……….(ii)

[Angle opposite to longer side is longer]

Adding eq. (i) and (ii),

4 + 3 <1 + 2C <A

(ii) In ABD, AB is the smallest side.

5 <8 ……….(iii)

[Angle opposite to smaller side is smaller]

In BDC, DC is the longest side.

6 <7 ……….(iv)

[Angle opposite to longer side is longer]

Adding eq. (iii) and (iv),

5 + 6 <7 + 8D <B

8. In figure, PR > PQ and PS bisects QPR. Prove that PSR>PSQ.

Ans. In PQR,PR > PQ[Given]

PQR>PRQ…..(i)[Angle opposite to longer side is greater]

Again1 = 2…..(ii)[ PS is the bisector of P]

PQR + 1 >PRQ + 2……….(iii)

ButPQS + 1 + PSQ = PRS + 2 + PSR = [Angle sum property]

PQR + 1 + PSQ = PRQ + 2 + PSR………(iv)

[PRS = PRQ and PQS = PQR]

From eq. (iii) and (iv),

PSQ<PSR

OrPSR>PSQ

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

Ans. Given: is a line and P is point not lying on PM

N is any point on other than M.

To prove: PM <PN

Proof: In PMN, M is the right angle.

N is an acute angle. (Angle sum property of )

M >N

PN> PM[Side opposite greater angle]

PM <PN

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

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

Ans. Let ABC be a triangle.

Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC. Let PQ bisects AB at M and RS bisects BC at point N.

Let PQ and RS intersect at point O.

Join OA, OB and OC.

Now in AOM and BOM,

AM = MB[By construction]

AMO = BMO = [By construction]

OM = OM[Common]

AOMBOM[By SAS congruency]

OA = OB[By C.P.C.T.]…..(i)

SimilarlyBON CON

OB = OC[By C.P.C.T.]…..(ii)

From eq. (i) and (ii),

OA = OB = OC

Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.

11. In a huge park, people are concentrated at three points (See figure).

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 ice cream parlour be set up so that maximum number of persons can approach it?

Ans. The parlour should be equidistant from A, B and C.

For this let we draw perpendicular bisector say of line joining points B and C also draw perpendicular bisector say of line joining points A and C.

Let and intersect each other at point O.

Now point O is equidistant from points A, B and C.

Join OA, OB and OC.

Proof: In BOP and COP,

OP = OP[Common]

OPB = OPC =

BP = PC[P is the mid-point of BC]

BOP COP[By SAS congruency]

OB = OC[By C.P.C.T.]…..(i)

Similarly, AOQCOQ

OA = OC[By C.P.C.T.]…..(ii)

From eq. (i) and (ii),

OA = OB = OC

Therefore, ice cream parlour should be set up at point O, the point of intersection of perpendicular bisectors of any two sides out of three formed by joining these points.

12. If ABC, the bisector of ABC and BCA intersect each other at the point O prove that BOC =

Ans.

Substituting this value of in (1)

So,

13. Prove that if one angle of a triangle is equal to the sum of the other two angles, the triangle is right angled:

Ans.Sum of three angles of triangle is ] …..(1)

Given that:

From (1) and (2)

Hence ABC is right angled.

14. IF fig, if PQ PS, PQSR, SQR = and QRT =, then find the values of X and Y.

Ans. is the transversal,

[pair of alternate angles]

Also in

15. If in fig, AD= AE and D and E are point on BC such that BD=EC prove that AB=AC.

Ans.

[Given]

[angles opposite to equal side are equal]

Now, [linear pair]

Also, [linear pair]

But,

Now in,

BD=CE

AD=AE

[By SAS]

[CPCT]

16. In the given figure, AC=BC, DCA=ECB and DBC=EAC. Prove that DBC andEAC are congruent and hence DC=EC.

Ans. We have,

[Given]

[adding $angle$ ECD on both sides]

...(i)

[From (i)

Now, in

[given]

[By SAS]

[CPCT]

17. From the following figure, prove that BAD=3ADB.

Ans.

Exterior = 2Q

Hence

18. O is the mid-point of AB and CD. Prove that AC=BD and ACBD.

Ans.

[O is the mid – point of AB]

[vertically opposite angles]

[O is the mid-point of CD]

[By SAS]

[CPCT]

Now, AC and BD are two lines inter sected by a transversal AB such that i.e. alternate angle are equal.

19. ABCD is a quadrilateral in which AD=BC and DAB=CBA. Prove that.

(i) ABDBAC

(ii) BA=AC

(iii) ABD=BAC

Ans.

[given]

[common]

[SAS criterion]

[CPCT]

20. AB is a line segment. AX and BY are equal two equal line segments drawn on opposite side of line AB such that AX BY. If AB and XY intersect each other at P. prove that

(i) APX BPY,

(ii) AB and XY bisect each other at P.

Ans.

[alternate angle]

[vertically opposite angle]

AX=BY[given]

[By AAS]

[CPCT]

AB and XY bisects each other at P.

21. In an isosceles ABC, with AB =AC, the bisector of B and C intersect each other at o, join A to o. show that:

(i) OB=OC

(ii) AO bisects A.

Ans. (i)
[given]
[angles opposite to equal side]

[side opposite to equal angle]

(ii)
[given]
[Halves of equals]
OB=OC [proved]
[SAS rule]
[CPCT]
i.e. AO bisects

22. Two side AB and BC and median AM of a triangle ABC are respectively equal to side PQ and QR and median PN of PQR, show that

(i)

(ii)

Ans. (i)
AB=PQ [Given]
BM=QN [Halves of equal]
AP=PN[Given]
[SSS rules]

(ii)
Now, in
AB=PQ [Given]
BC=QR [Given]

[SAS rule]

23. In the given figure, ABC and DBC are two triangles on the same base BC such that AB=AC and DB=DC. Prove that ABD =ACD,

Ans.

AB=AC[Given]

[angles opposite to equal side are equals]

Similarly in, [Given]…..(1)

……(2)

Adding (1) and (2)

24. Prove that the Angle opposite to the greatest side of a triangle is greater than two- third of a right angle i.e. greater than

Ans.

[angle opposite to large side is greater]….(i)
Similarly,
AB>AC

Adding (i) and (ii)

Adding to both sides,

[Sum of three angles of ]

25. AD is the bisector of A of ABC, where D lies on BC. Prove that AB>BD and AC>CD.

Ans.
[Exterior angles of is greater than each of the interior opposite angles]

But [Ad bisects ]
[Side opposite to greater angle is larger]

[Exterior angles of is greater than each of the interior opposite angles]
But,

[Side opposite to greater angle is larger].