Question 1

Question 1In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that Δ ABC ≅ Δ ABD. What can you say about BC and BD?

Accepted Answer

Given :AC = ADAB bisects ∠A i.e, ∠CAB = ∠DABIn Δ ABC and Δ ABD⇒ AB = AB (Common side)⇒ ∠CAB = ∠DAB (Proved above)⇒ AC = AD (Given)∴ Δ ABC ≅ Δ ABD (By S.A.S. Congruence rule)⇒ BC = BD (By C.P.C.T.)Hence, BC and BD are of equal length.

Question 2

Question 2ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that(i) Δ ABD ≅ Δ BAC(ii) BD = AC(iii) ∠ABD = ∠BAC

Accepted Answer

Given :AD = BC and ∠DAB = ∠CBA(i) In △ ABD and △ BAC,⇒ AD = BC (Given)⇒ ∠DAB = ∠CBA (Given)⇒ AB = BA (Common side)∴ △ ABD ≅ △ BAC (By S.A.S. congruence rule)Hence, proved that △ ABD ≅ △ BAC.(ii) As,△ ABD ≅ △ BAC,We know that,Corresponding parts of congruent triangles are equal.∴ BD = AC (By C.P.C.T.)Hence, proved that BD = AC.(iii) As,△ ABD ≅ △ BAC,We know that,Corresponding parts of congruent triangles are equal.⇒ ∠ABD = ∠BAC (By C.P.C.T.)Hence, proved that ∠ABD = ∠BAC.

Question 3

Question 3AD and BC are equal, perpendiculars to a line segment AB. Show that CD bisects AB.

Accepted Answer

Given :⇒ AD = BCFrom figure,⇒ AD ⊥ AB, ∠OAD = 90°⇒ BC ⊥ AB, ∠OBC = 90°In △ BOC and △ AOD,⇒ ∠BOC = ∠AOD (Vertically opposite angles are equal)⇒ ∠OBC = ∠OAD (Each equal to 90°)⇒ BC = AD (Given)∴ △ BOC ≅ △ AOD (By A.A.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.∴ BO = AO (By C.P.C.T.)Hence, proved that CD bisects AB and O is the mid-point of AB.

Question 4

Question 4l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that Δ ABC ≅ Δ CDA.

Accepted Answer

Given :l || m and p || qIn Δ ABC and Δ CDA,⇒ ∠BAC = ∠DCA (Alternate interior angles are equal)⇒ AC = CA (Common side)⇒ ∠BCA = ∠DAC (Alternate interior angles are equal)∴ Δ ABC ≅ Δ CDA (By A.S.A. congruence rule)Hence, proved that Δ ABC ≅ Δ CDA.

Question 5

Question 5Line 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. Show that:(i) Δ APB ≅ Δ AQB(ii) BP = BQ or B is equidistant from the arms of ∠A.

Accepted Answer

Given :l is the bisector of an angle ∠A and BP ⊥ AP and BQ ⊥ AQ(i) In Δ APB and Δ AQB,⇒ ∠BAP = ∠BAQ (l is the angle bisector of ∠A)⇒ ∠APB = ∠AQB (Each equal to 90°)⇒ AB = AB (Common side)∴ Δ APB ≅ Δ AQB (By A.A.S. congruence rule)Hence, proved that Δ APB ≅ Δ AQB.(ii) As,Δ APB ≅ Δ AQBWe know that,Corresponding parts of congruent triangles are equal.∴ BP = BQ (By C.P.C.T.)Hence, proved that BP = BQ or point B is equidistant from the arms of ∠A.

Question 6

Question 6In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Accepted Answer

Given :AC = AE, AB = AD,⇒ ∠BAD = ∠EAC.Adding ∠DAC to both sides of this equation, we get :⇒ ∠BAD + ∠DAC = ∠EAC + ∠DAC⇒ ∠BAC = ∠DAE.In Δ BAC and Δ DAE,⇒ AB = AD (Given)⇒ ∠BAC = ∠DAE (Proved above)⇒ AC = AE (Given)∴ Δ BAC ≅ Δ DAE (By S.A.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.∴ BC = DE (By C.P.C.T.)Hence, proved that BC = DE.

Question 7

Question 7AB 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. Show that(i) Δ DAP ≅ Δ EBP(ii) AD = BE

Accepted Answer

Given :P is the mid-point of AB.∴ AP = BP ......(1)Given,⇒ ∠BAD = ∠ABE .....(2)From figure,⇒ ∠BAD = ∠PAD and ∠ABE = ∠PBESubstituting values in equation (2), we get :⇒ ∠PAD = ∠PBE ........(3)(i) Given,⇒ ∠EPA = ∠DPB .........(4)Adding ∠DPE to both sides of the above equation,⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE∴ ∠DPA = ∠EPB ......(5)In Δ DAP and Δ EBP,⇒ ∠PAD = ∠PBE [From (3)]⇒ AP = BP [From (1)]⇒ ∠DPA = ∠EPB [From (5)]∴ Δ DAP ≅ Δ EBP (By A.S.A. congruence rule)Hence, proved that Δ DAP ≅ Δ EBP.(ii) As,Δ DAP ≅ Δ EBPWe know that,Corresponding parts of congruent triangles are equal.∴ AD = BE (By C.P.C.T.)Hence, proved that AD = BE.

Question 8

Question 8In 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. Show that :(i) Δ AMC ≅ Δ BMD(ii) ∠DBC is a right angle(iii) Δ DBC ≅ Δ ACB(iv) CM = 12AB\dfrac{1}{2}AB21​AB

Accepted Answer

(i) In Δ AMC and Δ BMD,⇒ AM = BM (M is the mid - point of AB)⇒ ∠AMC = ∠BMD (Vertically opposite angles are equal)⇒ CM = DM (Given)∴ Δ AMC ≅ Δ BMD (By S.A.S. congruence rule)Hence, proved that Δ AMC ≅ Δ BMD.(ii) Since,Δ AMC ≅ Δ BMD∴ ∠ACM = ∠BDM (By C.P.C.T.)From figure,∠ACM and ∠BDM are alternate interior angles. Since alternate angles are equal, it can be said that DB || AC.We know that,Sum of co-interior angles = 180°.⇒ ∠DBC + ∠ACB = 180°⇒ ∠DBC + 90° = 180° [Since, ΔACB is right angled triangle at point C]⇒ ∠DBC = 180° - 90°∴ ∠DBC = 90°.Hence, proved that ∠DBC is a right angle.(iii) Since,Δ AMC ≅ Δ BMD∴ DB = AC (By C.P.C.T.)In Δ DBC and Δ ACB,⇒ DB = AC (Proved above)⇒ ∠DBC = ∠ACB (Both equal to 90°)⇒ BC = CB (Common)∴ Δ DBC ≅ Δ ACB (By S.A.S. congruence rule)Hence, proved that Δ DBC ≅ Δ ACB.(iv) Since,Δ DBC ≅ Δ ACB⇒ AB = DC (By C.P.C.T.)⇒ 12AB\dfrac{1}{2}AB21​AB = 12DC\dfrac{1}{2}DC21​DCIt is given that M is the midpoint of DC⇒ CM = 12DC\dfrac{1}{2}DC21​DC = 12AB\dfrac{1}{2}AB21​AB∴ CM = 12AB\dfrac{1}{2}AB21​ABHence, proved that CM = 12AB\dfrac{1}{2}AB21​AB.

Question 9

Question 1In 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

Accepted Answer

Given :AB = ACOB is the bisectors of ∠B⇒ ∠ABO = ∠OBC = 12∠B\dfrac{1}{2}∠B21​∠B.OC is the bisectors of ∠C⇒ ∠ACO = ∠OCB = 12∠C\dfrac{1}{2}∠C21​∠C.(i) It is given that in triangle ABC, AB = AC⇒ ∠ACB = ∠ABCDividing both sides of equation by 2, we get :⇒∠ACB2=∠ABC2\Rightarrow \dfrac{∠ACB}{2} = \dfrac{∠ABC}{2}⇒2∠ACB​=2∠ABC​⇒ ∠OCB = ∠OBCWe know that,Sides opposite to equal angles of a triangle are also equal.⇒ OB = OC.Hence, proved that OB = OC.(ii) In Δ OAB and Δ OAC,⇒ AO = AO (Common)⇒ OB = OC (Proved above)⇒ AB = AC (Proved above)∴ Δ OAB ≅ Δ OAC (By S.S.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.⇒ ∠BAO = ∠CAO (By C.P.C.T.)∴ AO bisects ∠AHence, proved that AO bisects ∠A.

Question 10

Question 2In Δ ABC, AD is the perpendicular bisector of BC. Show that Δ ABC is an isosceles triangle in which AB = AC.

Accepted Answer

Given :AD is the perpendicular bisector of BC.∴ ∠ADB = ∠ADC = 90° and BD = DC.In Δ ADC and Δ ADB,⇒ AD = AD (Common side)⇒ ∠ADC = ∠ADB (Each equal to 90°)⇒ CD = BD (AD is the perpendicular bisector of BC)∴ Δ ADC ≅ Δ ADB (By S.A.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.∴ AB = AC (By C.P.C.T.)Hence, proved that ABC is an isosceles triangle in which AB = AC.

Question 11

Question 3ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively see Fig. Show that these altitudes are equal.

Accepted Answer

Given :Δ ABC is an isosceles triangle with AB and AC as equal sides.In Δ AEB and Δ AFC,⇒ ∠AEB = ∠AFC (Each equal to 90° as BE and CF are altitudes)⇒ ∠A = ∠A (Common angle)⇒ AB = AC∴ Δ AEB ≅ Δ AFC (By A.A.S. congruence rule)We know that,Corresponding parts of congruent triangles are equal.∴ BE = CF (By C.P.C.T.)Hence, proved that BE = CF.

Question 12

Question 4ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that(i) Δ ABE ≅ Δ ACF(ii) AB = AC, i.e., ABC is an isosceles triangle

Accepted Answer

Given :BE = CF, where BE and CF are altitudesSo, ∠AEB = 90° and ∠AFC = 90°(i) In Δ ABE and Δ ACF,⇒ ∠AEB = ∠AFC (Each 90°)⇒ ∠A = ∠A (Common angle)⇒ BE = CF (Given)∴ Δ ABE ≅ Δ ACF (By A.A.S. congruence rule)Hence, proved that Δ ABE ≅ Δ ACF.(ii) As,Δ ABE ≅ Δ ACFWe know that,Corresponding parts of congruent triangles are equal.∴ AB = AC (By C.P.C.T.)Hence, proved that ABC is an isosceles triangle with AB = AC.

Question 13

Question 5ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Accepted Answer

Given :ABC and DBC are isosceles triangles.Join AD.In △ DAB and △ DAC,⇒ AB = AC (Given)⇒ BD = CD (Given)⇒ AD = AD (Common side)∴ △ ABD ≅ △ ACD (By S.S.S. congruence rule)We know that,Corresponding sides of congruent triangles are equal.∴ ∠ABD = ∠ACD (By C.P.C.T.)Hence, proved that ∠ABD = ∠ACD.

Question 14

Question 6Δ 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.

Accepted Answer

Given :⇒ AB = AC .......(1)⇒ AD = AB ......(2)From equation (1) and (2), we get :⇒ AB = AC = ADIn an isosceles triangle ABC,⇒ AB = ACWe know that,Angles opposite to equal sides of a triangle are equal.∴ ∠ACB = ∠ABC = x (let)In Δ ACD,⇒ AC = ADWe know that,Angles opposite to equal sides of a triangle are equal.∴ ∠ADC = ∠ACD = y (let)From figure,⇒ ∠BCD = ∠ACB + ∠ACD⇒ ∠BCD = x + y .....(3)In Δ BCD,⇒ ∠ABC + ∠BCD + ∠ADC = 180° (Angle sum property of a triangle)⇒ x + (x + y) + y = 180° [From equation (1), (2) and (3)]⇒ 2(x + y) = 180°Substituting value of (x + y) from equation (3) in above equation :⇒ 2(∠BCD) = 180°⇒ ∠BCD = 180°2\dfrac{180°}{2}2180°​⇒ ∠BCD = 90°Hence, proved that ∠BCD is a right angle.

Question 15

Question 7ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Accepted Answer

Given :AB = AC and ∠A = 90°We know that,Angles opposite to equal sides are also equal.∠C = ∠B = x (let)In Δ ABC,⇒ ∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)⇒ 90° + ∠B + ∠C = 180°⇒ 90° + x + x = 180° (From(1))⇒ 2x = 180° - 90°⇒ 2x = 90°⇒ x = 90°2\dfrac{90°}{2}290°​⇒ x = 45°.∴ ∠B = ∠C = 45°Hence, ∠B = 45° and ∠C = 45°.

Question 16

Question 8Show that the angles of an equilateral triangle are 60° each.

Accepted Answer

Construct a triangle ABC with AB = BC = AC.In △ ABC,AB = BC = ACWe know that,Angles opposite to equal sides of a triangle are equal.∴ ∠C = ∠A = ∠B = x (let)In △ ABC,⇒ ∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)⇒ x + x + x = 180°⇒ 3x = 180°⇒ x = 180°3\dfrac{180°}{3}3180°​⇒ x = 60°.∴ ∠A = ∠B = ∠C = 60°.Hence, proved that the angles of an equilateral triangle are 60° each.

Question 17

Question 1Δ 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. 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

Accepted Answer

Given :Δ ABC and Δ DBC are isosceles triangles on the same base BC.∴ AB = AC and DB = DC(i) In Δ ABD and Δ ACD,⇒ AB = AC (Equal sides of isosceles Δ ABC)⇒ BD = CD (Equal sides of isosceles Δ DBC)⇒ AD = AD (Common)∴ Δ ABD ≅ Δ ACD (By S.S.S. congruence rule)Hence, proved that Δ ABD ≅ Δ ACD.(ii) Since,Δ ABD ≅ Δ ACD.We know that,Corresponding parts of congruent triangles are equal.⇒ ∠BAD = ∠CAD ........(1)From figure,⇒ ∠BAD = ∠BAP and ∠CAD = ∠CAPSubstituting above value in equation (1), we get :⇒ ∠BAP = ∠CAP ..........(2)In Δ ABP and Δ ACP,⇒ AB = AC (Equal sides of isosceles Δ ABC)⇒ ∠BAP = ∠CAP [From equation (2)]⇒ AP = AP (Common)∴ Δ ABP ≅ Δ ACP (By S.A.S. congruence rule)Hence, proved that Δ ABP ≅ Δ ACP.(iii) Since,Δ ABD ≅ Δ ACD∴ ∠ADB = ∠ADC (By C.P.C.T.) ..........(3)∴ ∠BAP = ∠CAP [From equation (2)]∴ AP is the angle bisector of ∠A.From equation (3),⇒ ∠ADB = ∠ADC⇒ 180° - ∠ADB = 180° - ∠ADC⇒ ∠BDP = ∠CDP ......(4)∴ AP is the bisector of ∠DHence, proved that AP bisects ∠A as well as ∠D.(iv) In Δ BDP and Δ CDP,⇒ DP = DP (Common side)⇒ ∠BDP = ∠CDP [From equation (4)]⇒ DB = DC (Equal sides of isosceles Δ DBC)∴ Δ BDP ≅ Δ CDP (By S.A.S. congruence rule)∴ ∠BPD = ∠CPD (By C.P.C.T.) .......(5)From figure,⇒ ∠BPD + ∠CPD = 180° (Linear pair)⇒ ∠BPD + ∠BPD = 180° [From Equation (5)]⇒ 2∠BPD = 180°⇒ ∠BPD = 90° .......(6)We know that,⇒ BP = CP [Proved above]Hence, proved that AP is the perpendicular bisector of BC.

Question 18

Question 2AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that(i) AD bisects BC(ii) AD bisects ∠A

Accepted Answer

Given :Δ ABC is an isosceles triangle and AB = AC.AD is altitude∴ ∠ADB = ∠ADC = 90°.(i) In Δ BAD and Δ CAD,⇒ ∠ADB = ∠ADC (Each equal to 90° as AD is altitude)⇒ AB = AC (Given)⇒ AD = AD (Common)∴ Δ BAD ≅ Δ CAD (By R.H.S. Congruence rule)We know that,Corresponding parts of congruent triangles are equal.∴ BD = CD (By C.P.C.T.)Hence, proved that AD bisects BC.(ii) Since, Δ BAD ≅ Δ CAD∴ ∠BAD = ∠CAD (By C.P.C.T.)Hence, proved that AD bisects ∠A.

Question 19

Question 3Two 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. Show that :(i) Δ ABM ≅ Δ PQN(ii) Δ ABC ≅ Δ PQR

Accepted Answer

Given :AB = PQ, BC = QR = x (let) and AM = PN.(i) Given,AM is the median of △ ABC∴ BM = CM = 12BC=x2\dfrac{1}{2}BC = \dfrac{x}{2}21​BC=2x​ .....(1)Also, PN is the median of △ PQR∴ QN = RN = 12QR=x2\dfrac{1}{2}QR = \dfrac{x}{2}21​QR=2x​ ......(2)From equation (1) and (2), we get :BM = QN ..........(3)Now, in △ ABM and △ PQN we have,⇒ AB = PQ (Given)⇒ BM = QN [From equation (3)]⇒ AM = PN (Given)∴ △ ABM ≅ △ PQN (By S.S.S. congruence rule)Hence, proved that △ ABM ≅ △ PQN.(ii) Since,△ ABM ≅ △ PQNWe know that,Corresponding parts of the congruent triangle are equal.∠B = ∠Q (By C.P.C.T.) ...........(4)Now, In △ ABC and △ PQR we have⇒ AB = PQ (Given)⇒ ∠B = ∠Q [From equation (4)]⇒ BC = QR (Given)∴ △ ABC ≅ △ PQR (By S.A.S. congruence rule)Hence, proved that △ ABC ≅ △ PQR.

Question 20

Question 4BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Accepted Answer

Triangle ABC with BE and CF as equal altitudes is shown in the figure below:Given :BE is a altitude.∴ ∠AEB = ∠CEB = 90°CF is a altitude.∴ ∠AFC = ∠BFC = 90°Also, BE = CF.In Δ BEC and Δ CFB,⇒ ∠BEC = ∠CFB (Each equal to 90°)⇒ BC = CB (Common)⇒ BE = CF (Given)⇒ Δ BEC ≅ Δ CFB (By R.H.S. congruence rule)We know that,Corresponding parts of congruent triangle are equal.⇒ ∠BCE = ∠CBF (By C.P.C.T.)As,Sides opposite to equal angles of a triangle are equal.∴ AB = AC.Hence, proved that Δ ABC is an isosceles triangle.

Triangles

Exercise 71

Exercise 72

Exercise 73