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Q1. Prove that the sum of any two sides of a triangle is greater than twice the length of median drawn to the third side.

Solution

Given: In triangle ABC, AD is the median drawn from A to BC. To prove: AB + AC > AD Construction: Produce AD to E so that DE = AD, Join BE. Proof: Now in ADC and EDB, AD = DE (by const) DC = BD(as D is mid-point) ADC = EDB (vertically opp. s) _{Therefore, In} ABE, ADC EDB(by SAS) This gives, BE = AC. AB + BE > AE AB + AC > 2AD ( AD = DE and BE = AC) Hence the sum of any two sides of a triangle is greater than the median drawn to the third side.

Q2. If then which of the following is true:

• 1) AC = RQ
• 2) AB = RP
• 3) CA = RP
• 4) CB = QP

Solution

By CPCT, CA = RP

Q3. In figure, if AB = AD, x = w and y = z then prove that AP = AQ.

Solution

Given: ABCD is a quadrilateral; Q and P are point on DC and BC respectively. To prove: AP = AQ Proof : Since, w = x and y = z _{Adding, we get} w + z = x + y DAC = BAC In ADC and BAC In QAC and PAC Hence AP = AQ.

Q4. Prove that diagonals of a rectangle are equal but not perpendicular.

Solution

Q5. Two equilateral triangles are congruent when:

• 1) Their sides are equal
• 2) Their angles are equal
• 3) Their sides are proportional
• 4) Their areas are proportional

Solution

Their sides are equal

Q6. In figure, AC = AE, AB = AD and BAD = EAC show that BC = DE.

Solution

_Given: AC = AE, AB = AD and BAD = EAC To prove: BC = DE _Proof:BAD = EAC (given) BAD + DAC = EAC + DAC BAC = DAE Now ABC and ADE

Q7. Given OAP OBP in below figure, the criteria by which the triangles are congruent:

• 1) ASA
• 2) SSS
• 3) RHS
• 4) SAS

Solution

Since two sides and included angles are equal, the criteria is SAS.

Q8.

• 1) QR= 5 cm, R= 60⁰
• 2) QP = 5 cm, P= 60⁰
• 3) QR= 5 cm, Q= 40⁰
• 4) QP= 5 cm, R = 60⁰

Solution

Q9. In fig., it is given that ABD BAC. Name the criteria by which the triangles are congruent.

• 1) SAS
• 2) RHS
• 3) ASA
• 4) SSS

Solution

The triangles are congruent by RHS criterion.

Q10. In figure, D and E are points on the base BC of a ABC such that BD = CE and AD = AE. Prove that ABE ACD.

Solution

In ACD and ABE AD = AE (given) BD = EC (given) BD + DE = DE + EC BE = CD ADC = AEB (AD = AE)

Q11. In ABC, AB = AC. D is point inside ABC such that BD = DC. Prove that ABD = ACD.

Solution

Here, AB = AC ABC = ACB (Angles opp. to equal sides are equal) Also, BD = DC DBC = DCB (Angles opp. to equal sides are equal) _Subtracting, ABC - DBC = ACB - DCB ABD = ACD

Q12. A point O is taken inside an equilateral four sided figure ABCD such that its distances from the angular points D and B are equal. Show that AO and OC are in one and the same straight line.

Solution

In AOD and AOB AD = AB                                             (given) AO =AO                                             (common side) OD = OB                                             (given) AOD  AOB                          (SSS congruence rule)  AOD = AOB                          (c.p.c.t) Similarly, DOC BOC                (SSS congruence rule) DOC = BOC                           (c.p.c.t) AOD + AOB + DOC + BOC = 360^o         (angles at a point) 2AOD + 2DOC = 360^o AOD + DOC = 180^o Hence, AO and OC are in one and the same straight line.

Q13. In fig., if AC = BC, DCA = ECB and DBC = EAC then Prove that BD = AE.

Solution

Given that … (i) Now in DBC and EAC [from (i)] BC = AC (Given) (Given) (ASA Congruence) Therefore, BD =AE (CPCT)

Q14. In ∆ABC, AB = AC, AD = AE, ∠BAD = ∠CAE, then  

• 1) Ar(∆ABE) = Ar(∆ACD)
• 2) Insufficient information
• 3) Both A and B
• 4) ∆ABD ≅ ∆ACE

Solution

In ∆ABD and ∆ACE, AB = AC  ∴ ∠ABD = ∠ACE Opposite angles of equal sides ∠BAD = ∠CAE ∆ABD ≅ ∆ACE ASA Ar(∆ABD) = Ar(∆ACE) Ar(∆ABD) + ar(∆ADE) = Ar(∆ACE) + ar(∆ADE) Ar(∆ABE) = ar(∆ACD)

Q15. In ΔABC if AB = BC then:

• 1) A = B
• 2) B > C
• 3) A < C
• 4) A = C

Solution

Q16. If length of the largest side of a triangle is 12 cm then other two sides of triangle can be:

• 1) 7.6 cm, 3.4 cm
• 2) 4.8 cm, 8.2 cm
• 3) 3.2 cm, 7.8 cm
• 4) 6.4 cm, 2.8 cm

Solution

Since, in a triangle, the sum of two sides is greater than the third side therefore other two sides are: 4.8 cm, 8.2 cm.

Q17. In the figure PQ > PR, QS and RS are the bisectors of Q and R respectively. Prove that SQ > SR.

Solution

Q18. In figure, ABC is an isosceles triangle with AB = AC, BD and CE are two medians of the triangle, prove that BD = CE.

Solution

In EBC and DCB, AB = AC (given) i.e., BE = CD C = B ( AB = AC) BC = BC (common) Thus, CE = BD (By c.p.c.t.)

Q19. In fig., A is a point equidistant from two lines l₁ and l₂ intersecting at a point P. Show that AP bisects the angle between l₁ and l₂.

Solution

Given: A point A equidistant from 2 lines intersecting at a point P. To prove: AP bisects angle between Proof: In APM and APN AM = AN (Given) AMP = ANP = 90^o (Given) AP=AP (Common) (RHS congruency rule) APM = APN

Q20.

Solution

Q21. Prove that ABC is an isosceles triangle if median AP is perpendicular to the base BC:

Solution

Q22. In ABC, if AB > BC then:

• 1) A = B
• 2) C < A
• 3) C =A
• 4) C> A

Solution

_{Angle opp. to bigger side is greater. Therefore,} C> A

Q23. Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle.

Solution

Given: Two triangles ABC and DEF such that B = E, C = F and BC = EF. To prove: ABC DEF Proof: Case I: If AB = DE then in ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given] Thus, ABC DEF [SAS criterion] Case II: If AB < DE Take a point G on ED such that EG = AB. Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion] ACB = GFE [corresponding parts of congruent triangles are equal] But ACB = DFE [given] GFE = DFE, This is only possible when FG coincides with FD or G coincides with D. AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED With a similar argument (as in case II), we may conclude that ABC DEF (by SAS) Thus, ABC DEF.

Q24. In figure, AC= AE, AB= AD and BAD = EAC. Prove that BC= DE.

Solution

Join DE --------- (i) BC = DE (CPCT)

Q25.

Solution

Q26. In figure, D is the mid-point of base BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that B = C.

Solution

In BED and CFD BED = DFC = 90^o BD = DC (D is the mid-point of BC) ED = FD (given) B = C (By c.p.c.t)

Q27. In figure, AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that 2APD = B + C.

Solution

APD = (B +C) 2[APD] = B + C

Q28. If AD is the bisector of A of ABC, show that AB > DB.

Solution

Given: AD is the bisector of A To prove: AB > DB. Proof : Since AD is bisector, But 3 > 2 3 > 1 So, AB > DB (side opp. to greater angle is larger).

Q29. In figure, AB = BC, AD = EC. Prove that ABE CBD.

Solution

AD = EC (Given) Adding DE both side AD + DE = EC + DE AE = CD In ABE and DBC BA = BC (given) A = C (angles opposite to equal sides are equal) AE = CD (by (i)) ABE DBC [by SAS]

Q30. In figure, D is a point on side BC of such that AD=AC. Show AB>AD.

Solution

Q31. In figure, QPR = PQR and M and N are respectively points on sides QR and PR of PQR, such that QM = PN. Prove that OP = OQ, where O is the point of intersection of PM and QN.

Solution

Q32. ABCD is a quadrilateral in which AB = AD, BC = DC. AC and BD intersect at E. Prove that, AC bisects each of the angles A and C.

Solution

Q33. In ABC, D is the midpoint of BC. The perpendiculars from D to AB and AC are equal. Prove that ABC is isosceles.

Solution

Given: D is the midpoint of BC and PD = DQ To prove: ABC is isosceles Proof: In _{BDP and CDQ BD = DC (D is mid-point) PD = QD (given) BPD = CQD (each 90 degrees) Therefore, BDP CDQ (By RHS) So, BP = CQ … (1) (By CPCT)} Again in _{APD and AQD AD = AD (common) PD = QD (given) APD = AQD (each 90 degrees) Therefore, APD AQD (By RHS) So, PA = QA … (2) (By CPCT) Adding (1) and (2), we get BP + PA = CQ + QA Or, AB = AC.}

Q34. In the given figure, ABD = 130° and EAC = 120°. Prove that AB > AC.

Solution

130⁰+ABC = 180⁰[linear pair] ABC = 180^o-130^o = 50^o ….(i) In ABC, side BA is extended to E So EAC = ABC+ACB (exterior angle is equal to the sum of pair ofinterior opposite angles) ACB = 120⁰ - 50⁰ = 70⁰ InABC,ACB> ABC Thus, AB>AC [side opposite to greater angle is larger]

Q35. In fig., O is the mid-point of AB and CD. Prove that AC = BD.

Solution

OB= OA (O is the mid-point of AB) OC=OD (O is the mid-point of CD) By SAS rule,

Q36. In a ΔDEF, if D = 30° E = 60° then which side of the triangle is longest and which side is shortest?

Solution

In DEF, D + E + F = 180^o(Angle sum property) 30^o + 60^o +F = 180^o F = 90^o F = 90^o is the largest angleDE is longest side D = 30^o is the smallest angle EF is smallest side

Q37. In fig., if AB = AC and PB = QC, then by which congruence criterion PBCQCB

• 1) SAS
• 2) RHS
• 3) ASA
• 4) SSS

Solution

In PBC_and QCB PB = QC(given) B = C(as AB = AC) BC = BC(common) Therefore, PBCQCB(by SAS)

Q38. In triangle ABC, AD is a median. BL and CM are perpendiculars on the median (or the median extended). Prove that BL = CM.

Solution

Q39. In ABC, if A = 50^o and B = 60^o, then the greatest side is:

• 1) BC
• 2) AB
• 3) Cannot say
• 4) AC

Solution

Third C = 180^o - 50^o - 60^o = 70^o Since  C is largest, so the side opposite to C that is AB is greatest.

Q40. In fig., D is any point on the base BC produced of an isosceles triangle ABC. Prove that AD > AB.

Solution

Given: D is any point on the base BC produced of an isosceles triangle ABC. To prove: AD > AB Proof : AB = AC (Since, ABC is an isosceles triangle) ABC = ACB(angles opposite equal sides are equal) In ACD, ext ACB > ADC

Q41. In a triangle ABC, AB = 3 cm, BC = 3.5 cm and CA = 6.5 cm. Is it possible to construct this triangle? Give reason for your answer.

Solution

The construction of the triangle with the given sides is not possible as the sum of any two sides must be greater than the third side, which is not true in this case. Here, AB + BC = 6.5 AC = 6.5 So, we see that the required condition is not satisfied.

Q42. Two isosceles triangles are on opposite sides of a common base, then by which criterion can we sayABD ACD?

• 1) RHS
• 2) ASA
• 3) SSS
• 4) SAS

Solution

In the two triangles ABD and ACD, AB = AC,    BD = CD, AD = AD. ABD ACD by SSS

Q43. In right triangle ABC, C = 90°, M is 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. Show that (i) ΔAMC ΔBMD (ii) DBC = ACB.

Solution

(i) In AMC and BMD AM = BM (Given) AMC = BMD (Vertically opposite angle) MC = MD (given) By SAS congruency (ii)

Q44. In fig., if AB = AC and BD = DC. ABD and ACD are congruent by which criterion.

• 1) SAS
• 2) ASA
• 3) RHS
• 4) SSS

Solution

By SSS.

Q45. In the given figure, ABCD is a quadrilateral in which AB || DC and P is the mid point of BC. On producing, AP and DC meet at Q. Prove that(i) AB = CQ,(ii) DQ = DC + AB

Solution

Q46. In figure , AB > AC, BO and CO are the bisectors of B and C respectively. Show that OB > OC.

Solution

AB>AC (given) Therefore C > B[angle opposite to the longer side is greater] So C > B i.e. OCB > OBC Therefore OB > OC

Q47.

Solution

Q48.

Solution

Since the measure of angle Q is least so the side opposite to Q is smallest.

Q49. The side BC of ABC is produced on both sides. Prove that the sum of the two exterior angles so formed is greater than A by 180^o_.

Solution

Given: Side BC of ABC is produced on both sides. To prove: ext B + ext C > A Proof: ext B + ext C = 180^o -B + 180^o - C = 360^o - (B + C) = 360^o - (180^o - A) = 180^o + A Hence, ext B + ext C > A.

Q50. ΔABC is an isosceles triangle with AB = AC, side BA is produced to D such that AB = AD. Prove that BCD is right angle.

Solution

  is an isosceles triangle with AB = AC [given] Add (1) and (3)

Q51. In an isosceles ABC with AB = AC, BD and CE are two medians. Prove that BD = CE.

Solution

Given: In an isosceles ABC with AB = AC, BD and CE are two medians. To Prove: BD = CE Proof: In triangles ABD and ACE AB = AC (given) AD = AE (as D and E are the mid-points of AC and AB) A = A Therefore, triangles ABD and ACE are congruent by SAS. Hence, BD = CE by CPCT.

Q52. In the following figure, AD is a median of and BL and CM are perpendiculars drawn from B and C respectively on AD and AD produced respectively. Prove That BL=CM.

Solution

Q53. A triangle can have:

• 1) Two obtuse angles
• 2) Two right angles
• 3) All angles more than 60^o
• 4) Two acute angles

Solution

Two acute angles

Q54. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that BCD is a right angle.

Solution

Given: ABC is an isosceles triangle in which AB = BC BA is produced to D such that AB = AD To prove: BCD = 90^o Proof: In ABC AB = AC 1 = 2 … (i) Similarly, 3 = 4 … (ii) In BCD, by angle sum property, 1 + (2 + 3) + 4 = 180^o ₂ + (2 + 3) + 3 = 180^o 2(2 + 3) = 180^o 2 + 3 = 90^o BCD = 90^o

Q55. In ΔABC, AB = AC, A = 36°. If the internal bisector of C meet AB at D, prove that AD = BC.

Solution

  AB = AC C = B [angles opposite to equal sides are equal] In ABC A + B + C = 180^o [Angle sum property of ] 36^o   + 2C = 180^o 2C = 144^o _{C = 72}^o DCA = DCA = DAC = 36^o AD = CD [sides opposite to equal angles are equal] … (i) In CDB AB C + C + CDB = 180^o 72^o + 36^o + CDB = 180^o CDB = 180^o - 72^o - 36^o CDB = 72^o CDB = DBC = 72^o BC = CD [sides opposite to equal angles are equal] … (ii) AD = BC (Using (i) and (ii))

Q56. In ABC, AB = AC. X and Y are points on AB and AC such that AX = AY. Prove that ABY ACX

Solution

In ^s ABY and ACX, AB = AC (given) AX = AY (given) A = A (common) By SAS, ABY ACX

Q57. In fig., if a > b then prove that PQ > PR.

Solution

a = P + R[Exterior angle property] b = P + Q a > b given R > Q PQ > PR(sides opposite to larger angles are longer) Hence proved.

Q58. In an isosceles triangle 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 bisect A

Solution

Given: An isosceles triangle ABC with AB = AC. To prove : (1) OB = OC (2) AO bisector A Construction: Join AO Proof: ABC = ACB (given) So OB = OC (2) In AOB and AOC OB = OC AB = AC AO = AO Thus AOB ABC (By SSS) Hence, OAB = OAC (By CPCT)

Q59. In fig., PR > PQ and PS bisects QPR. Prove that PSR > PSQ.

Solution

In PQR PR>PQ (given) PQR>PRQ [Angle opp. to larger side is greater] PQR+1>PRQ+2 … (i) [1= 2 given] In PQS and PSR PQR+1+PSQ = 180⁰ and PRQ+2+PSR= 180⁰ PQR+1 = 180⁰-PSQ and PQR+2 = 180⁰-PSR 180⁰-PSQ>180⁰-PSRfrom (i) PSR-PSQ>0 PSR>PSQ

Q60. In figure , if OA = OB, OD= OC then by congruence rule:

• 1) ASA
• 2) SSS
• 3) SAS
• 4) RHS

Solution

also, OA = OB, OD= OC  SAS

Q61. It is not possible to construct a triangle when its sides are:

• 1) 8.3 cm, 3.4 cm, 6.1 cm
• 2) 3 cm, 5 cm, 5 cm
• 3) 6 cm, 7 cm, 10 cm
• 4) 5.4 cm, 2.2 cm, 3.1 cm

Solution

In a triangle, the sum of two sides is always greater than third side. But in option (B), 2.2 + 3.1 = 5.3 < 5.4.

Q62. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that AD is also the median of the triangle.

Solution

Given: AD is an altitude of an isosceles triangle ABC in which AB = AC To prove: AD is also the median of the triangle Proof: In ABD and ACD, AB = AC (given) ADB = ADC (each 90^o) AD = AD (common) ABD ACD (By SAS) BD = DC AD is the median.

Q63.

Solution

Q64. In ABC, the bisector AD of A is perpendicular to side BC. Show that ABC is an isosceles triangle.

Solution

(By C.PC.T) Hence, ABC is an isosceles triangle.

Q65.

Solution

Q66. It is given that AB = EF, BC = DE, AB  BD and FE  CE. Prove that ABD FEC.

Solution

Q67. In figure, AB = AC, D is the point in the interior of ABC such that DBC = DCB. Prove that AD bisects BAC of ABC.

Solution

AD = AD common Hence, AD is bisector of

Q68.

Solution

Q69.

Solution

Q70.

Solution

Q71. 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 the given figure). Show that: (i) AMC BMD (ii) DBC is a right angle. (iii) DBC ACB (iv) CM = AB

Solution

In AMC and BMD AM = BM (M is midpoint of AB) AMC = BMD (vertically opposite angles) CM = DM (given) AMC @ BMD (by SAS congruence rule) AC = BD (by CPCT) And ACM = BDM (by CPCT) (ii) We have ACM = BDM But ACM and BDM are alternate interior angles Since alternate angles are equal. Hence, we can say that DB || AC DBC + ACB = 180º (co-interior angles) DBC + 90º = 180º DBC = 90º (iii) Now in DBC and ACB DB = AC (Already proved) DBC = ACB (each 90) BC = CB (Common) DBC ACB (SAS congruence rule) (iv) We have DBC ACB AB = DC (by CPCT) AB = 2 CM CM =AB

Q72. If the altitudes from two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.

Solution

Q73. In triangle ABC, D is a point on AC, Prove that, the value of the perimeter of triangle ABC is greater than twice the length of BD.

Solution

Q74.

Solution

Q75. Prove that angles opposite to equal sides of a triangle are equal.

Solution

Given: In triangle ABC, AB = AC. To prove: _{B = C Construction: Draw AD perpendicular to BC. Proof: In ADB and ADC AB = AC(given) 1 = 2(each 90o) AD = AD(common) Therefore, In ADB ADC(By RHS) Thus, B = C(By cpct) Hence proved.}

Q76. In figure, two sides AB and BC and the median AM of ABC are respectively equal to sides DE and EF and the median DN of DEF. Prove that ABC DEF.

Solution

Given: AB = DE, BC = EF and AM = DN To prove: ABC DEF Proof: In ABM DEN AB = DE (given) AM = DN (given) BM = EN (as M and N are the respective mid-points and BC = EF) Therefore, ABM DEN (by SSS) Thus, _{B = E (By cpct) … (1) Again in} ABC _and DEF _{B = E by (1)} AB = DE (given) BC = EF (given) Therefore, ABC DEF (By SAS) Hence proved.

Q77. In figure, two isosceles triangles have a common base, prove that line segment joining their vertices bisects the common base at right angles.

Solution

In ABD and ACD, AB = AC (given) BD = DC (given) AD = AD (common) Therefore, ABD ACD (By SSS rule) So, BAD = CAD (By cpct) ... (1) In ABE and ACE, AB = AC (given) AE = AE (common) BAD = DAC (By (1)) ...(2) Therefore, ABE ACE (By SAS) BE = CE (By cppt) And, BEA = CEA (cpcpt) But, BEA + CEA = 180^o BEA = 90^o

Q78.

Solution

Q79. In the given figure, x > y. Show that LM > LN.

Solution

LMN + x = 180^o                  (linear pair)     LNM + y = 180^o                  (linear pair)   LMN + x = LNM + y     But, it is given that  x > y.   LMN < LNM   LM > LN                            (side opposite greater angle is greater)

Q80. Two sides AB and BC and median AM of ABC are respectively equal to the sides PQ and QR and median PN of PQR. Show that ABM PQN.

Solution

Given :in which AB=PQ,BC=QR and median AM=PN To Prove: Proof BC = QR AB=PQ and AM=PN (Given) Therefore, (by SSS)

Q81. Show that in a right angled triangle the hypotenuse is the longest side.

Solution

Given :A right angledtriangle ABC, B = 90^o To Prove:Hypotenuse AC in the longest side. (i) AC > AB(ii)AC > BC Proof: Now in ABC ABC = 90^o But CAB + BCA +CBA = 180^o Thus BCA < 90^o and CAB < 90^o (Side opposite to greater angle is larger) Thus AC is the longest side

Q82. In ABC, BD and CE are two altitudes such that BD = CE. Prove that ABC is isosceles.

Solution

In ABD and ACE ADB= AEC= 90⁰ BAD= CAE (common angle) BD = CE (given) By AAS Criteria ABD ACE AB = AC (CPCT) Hence, ABC is isosceles.

Q83. In figure, AB and CD are respectively the smallest and the longest sides of a quadrilateral ABCD, then show that A > C and B > D.

Solution

Let us join AC. In DABC AB < BC (AB is smallest side of quadrilateral ABCD) 2 < 1 (angle opposite to smaller side is smaller) ... (1) In DADC AD < CD (CD is the largest side of quadrilateral ABCD) 4 < 3 (angle opposite to smaller side is smaller) ...(2) On adding equations (1) and (2), we have 2 + 4 < 1 + 3 C < A A > C   Let us join BD. In DABD AB < AD (AB is smallest side of quadrilateral ABCD) 8 < 5 (angle opposite to smaller side is smaller) ... (3) In DBDC BC < CD (CD is the largest side of quadrilateral ABCD) 7 < 6 (angle opposite to smaller side is smaller) ... (4) On adding equations (3) and (4), we have 8 + 7 < 5 + 6 D < B B > D

Q84. In figure, BCD = ADC and ACB = BDA. Prove that AD = BC and A = B.

Solution

1= 2 and 3 = 4 1+3 = 2+4 ACD = BDC In ACD and BDC ADC = BCD (given) CD = CD (COMMON) ACD = BDC [from (i)] ACD BDC (ASA rule) AD = BC and A = B (CPCT)

Q85. ABCD is a square, X and Y are the points on the sides AB and DC respectively such that BY = CX. Prove that CY = BX and CBY = BCX.

Solution

In the right triangles CBY and BCX BY = CX                                 (given) CB = BC                                 (common) YCB = XBC                   (90^o each) CBY BCX                    (RHS congruence rule)  CY = BX and CBY = BCX.

Q86. In fig., AD and BC are perpendiculars to a line segment AB such that AD = BC. Prove that CD bisects AB at P.

Solution

In BCP and ADP B = A = 90^o APD = BPC (vertically opp. angles) And BC = AD (given) BCD ADP (By AAS) CP = DP and AP = BP Hence, CD bisects AB at P.

Q87.

Solution

Q88.

Solution

Q89. Show that of all line segment drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution

    Let AB be a perpendicular to a line l and AP is any other line segment In right ABP , B > P (B = 90^o) _Therefore, AP > AB_{. (In a triangle, side opposite to larger angle is longer) Hence, the result.} 

Q90. In fig., ABC and DBC are two isosceles triangles on the same base BC and the vertices A and D are on the same side of BC. AD is extended to meet BC at P. Prove that AP bisects A. .

Solution

Given: Two isosceles triangles ABC and DBC. To Prove: AP bisects A. Proof: In ABD and ACD AB = AC (given) BD = CD (given) AD = AD (common) _{So by CPCT}BAD = CAD AP bisects BAC.

Q91. If the segments drawn perpendicular to the two sides of a triangle from the mid point of the third side be congruent and equally inclined to the third side, prove that the triangle is isosceles.

Solution

Q92. In figure, ABCD is a square and DEC is an equilateral triangle. Prove that: (i) ADE BCE (ii) AE = BE (iii) DAE = 15°

Solution

But

Q93. In fig., if ABD = ACE and AB = AC. Prove that BD = CE.

Solution

In ABD as ACE AB = AC (given) ABD = ACE (given) BAD = CAE (common angle) ABD ACE (By ASA) Thus, BD = CE (CPCT)

Q94.

Solution

Q95. In PQR, if PQ = QR and L, M and N are the mid points of the sides PQ, QR and RP respectively. Prove that LN = MN.

Solution

Q96. In quadrilateral PQRS the diagonals intersect at O than prove that PQ + QR + RS + PS > PR + QS.

Solution

Q97. In figure, show that 2 (AC + BD) > (AB + BC + CD + DA)

Solution

In OAB OA + OB > AB ...(1) In OBC OB + OC > BC...(2) In OCD, OC + OD > CD...(3) In ODA, OD + OA > AD ...(4) Adding (1), (2), (3) and (4) 2 (OA + OB + OC + OD) > AB + BC + DC + DA 2|(OA + OC) + (OB + OD)| > AB + BC + CD + DA 2 (AC+ BD) > AB + BC + CD + DA

Q98. In fig., ABCD is a quadrilateral with AC = AD and AB bisects A. Show that ΔABC ΔABD. What can you say about BC and BD?

Solution

In ABC and ABD AC = AD (given) CAB = DAB [AB is bisection of DAC] AB = AB (common) So, by SAS congruency ABC ABD BC = BD (CPCT)

Q99. ABCD is a parallelogram, if the two diagonals are equal, find the measure of ABC.

Solution

AB = CD and AD = BC           (opposite sides of a parallelogram are equal) In ABD and BAC AD = BC BD = AC                               (Given) AB = BA                                (Common) ABD BAC                (SSS congurence rule) BAD = ABC              (c.p.c.t) AD  BC and transversal AB intersects them at A and B. BAD +ABC = 180^o ABC +ABC = 180^o  2ABC = 180^o    ABC = 90^o Hence, the measure of ABC = 90^o

Q100. If two isosceles triangles have a common base, prove that the line segment joining their vertices bisects the common base at right angles.

Solution

Since AB = AC, BD = DC, AD = AD In ABP and ACP AB = AC, Hence, AP bisects common base at right angle.

Q101. In figure, if AB=AC and DB=DC then prove that

Solution

Q102. In figure, and O is the midpoint of AD. Show that (i) (ii) O is also the midpoint of BC.

Solution

_(i) In (Alternate angle as AB||DC and BC is transversal) (Vertically opposite angles) OA = OD (Given) _(ii) OB = OC (CPCT) So, O is midpoint of BC.