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EXERCISE 7.1 1. In a quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show

that DABC ≅ DABD. What can you say about BC and BD?

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

(i) DABD ≅ DBAC (ii) BD = AC (iii) ∠ABD = ∠BAC

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

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4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that DABC ≅ DCDA.

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 figure).

Show that: (i) DAPB ≅ DAQB (ii) BP = BQ or B is equidistant from the

arms of ∠A.

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

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 figure). Show that:

(i) DDAP ≅ DEBP (ii) AD = BE.

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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 figure). Show that:

(i) DAMC ≅ DBMD (ii) ∠DBC is a right angle (iii) DDBC ≅ DACB

(iv) CM = 12AB.

Test Yourself – TR 1 1. In the figure, it is given that AE = AD and BD = CE. Prove that DAEB ≅

DADC.

2. In the figure, PQRS is a quadrilateral and T and U are respectively points on PS and RS such that PQ = RQ, ∠PQT = ∠RQU and ∠TQS = ∠UQS. Prove that QT = QU.

3. In an isosceles triangle ABC, AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.

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4. In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that DCAP ≅ DBAP and hence CP = BP.

A

B

C

D

P

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

6. In the figure, it is given that ∠A = ∠C and AB = BC. Prove that DABD ≅ DCBE.

7. In the figure, AD is a median and BL, CM are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BL = CM.

8. If DABC is an isosceles triangle such that AB = AC, then prove that altitude AD from A on BC bisects BC.

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9. In the figure, it is given that AB = EF, BC = DE, AB ^ BD and FE ^ CE. Prove that: DABD ≅ DFEC.

10. In Ds ABC and PQR, AB = PQ, BC = QR and CB and RQ are extended to X and Y respectively and ∠ABX = ∠PQY. Prove that DABC ≅ DPQR.

EXERCISE 7.2 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

2. In DABC, AD is the perpendicular bisector of BC (see figure). Show that DABC is an isosceles triangle in which AB = AC.

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

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

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

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

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6. DABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

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

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

Test Yourself – TR 2 1. In the figure, AB = AC, ∠A = 48º and ∠ACD = 18º. Show that BC = CD.

2. Calculate angles a, b and c.

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3. In the figure, AB = AC. D is the mid-point of BC. Prove that

(i) DP = DQ (ii) AP = AQ (iii) AD bisects ∠A. 4. In the figure, it is given that AB = BC and AD = CE. Prove that DABE ≅

DCBD.

5. In the figure, ABC is an isosceles triangle in which AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE.

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

7. DABC is an isosceles triangle with AB = AC and BE, CF are the bisectors of angles B and C respectively. Prove that DEBC ≅ DFCB.

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8. Find ∠ADC in the given figure.

9. PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that: PS = PT.

10. In the following figure, PS = PR, ∠TPS = ∠QPR. Prove that: PT = PQ.

EXERCISE 7.3 1. DABC and DDBC 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) DABD ≅ DACD (ii) DABP ≅ DACP (iii) AP bisects ∠A as well as ∠D. (iv) AP is the perpendicular bisector of BC.

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

(i) AD bisects BC (ii) AD bisects ∠A.

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 DPQR (see figure). Show that:

(i) DABM ≅ DPQN (ii) DABC ≅ DPQR

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

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

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Test Yourself – TR 3 1. In the figure, BA ^ AC and DE ^ EF such that BA = DE and BF = DC.

Prove that AC = EF.

2. In the figure, if PQ = PR and OQ = OR. Find the ratio ∠PQO : ∠PRO.

3. AD, BE and CF are three altitudes of a triangle ABC, which are equal. Prove that DABC is an equilateral triangle.

4. In the figure, if DF = DE, BD = CD, DE ^ AC and DF ^ AB, prove that AB = AC.

5. The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown. Prove that the image is as far behind the mirror as the object is in-front of the mirror.

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6. The vertices of two Ds drawn on the opposite sides of a common base are equidistant from the base. Show that the base bisects the line joining the vertices.

7. In the following fig, equal sides BA and CA of DABC are produced to Q and P respectively so that AP = AQ. Prove that PB = QC.

8. In the following fig., BC = CE and ∠1 = ∠2 . Prove that DGCB ≅ DDCE.

9. In a DABC, the internal bisectors of ∠B and ∠C meet at O. Prove that OA is the internal bisector of angle A.

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10. In the given fig., DABC is right angled at B. ACDE and BCGF are squares. Prove that:

(i) DDCB ≅ DACG (ii) AG = DB

EXERCISE 7.4 1. Show that in a right angled triangle, the hypotenuse is the longest side.

2. In the figure, sides AB and AC of DABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

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3. In the figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

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

5. In the figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

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

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Test Yourself – TR 4 1. In a DABC, if ∠A = 55º and ∠B = 65º. Find the shortest and longest sides

of the triangle. 2. In the figure, PQ > PR, QS and RS are the bisectors of ∠Q and ∠R

respectively. Prove that SQ > SR.

3. If AD is the bisector of ∠A, show that AB > BD. 4. In the figure, AC > AB and D is the point on AC such that AB = AD. Prove

that CD > BC.

5. In the following fig., prove that: (i) MN + NO + OP + PM > 2 MO (ii) MN + NO + OP > PM

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6. In the following fig., PQ = PR, show that PS > PQ.

7. In the following fig., AB > AC. Show that AB > AD.

8. In the following fig., ∠E > ∠A and ∠C > ∠D. Prove that AD > EC.

9. Prove that any two sides of a triangle are together greater than twice the median drawn to the third.

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10. AB and CD intersect each other at O. Prove that: AM + BM+ CM + DM > AO+ BO + CO + DO

EXERCISE 7.5 (OPTIONAL) 1. ABC is a triangle. Locate a point in the interior of DABC which is equidistant

from all the vertices of DABC. 2. In a triangle locate a point in its interior which is equidistant from all the

sides of the triangle. 3. 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?

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4. Complete the hexagonal and star shaped Rangolies [see figure (i) and (ii)] by filling them with as many equilateral triangles of side 1cm as you can. Count the number of triangles in each case. Which has more triangles?

5 cm

5 cm

5 cm

5 cm

5 cm5 cm

(i) (ii)

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NCERT Exercises and Assignments

Exercise 7.1 1. Now, in Ds ABC and ABD, we have AC = AD (given) ∠CAB = ∠DAB (Q AB bisects ∠A) and AB = AB (common) \ By SAS congruence criterion, we have DABC ≅ DABD So, BC = BD (CPCT) 2. In Ds ABD and BAC, we have AD = BC (given) ∠DAB = ∠CBA (given) AB = BA (common) \ By SAS criterion of congruence, we have DABD ≅ DBAC, which proves (i) BD = AC and ∠ABD = ∠BAC, which proves (ii) and (iii) (CPCT) 3. In Ds AOD and BOC, we have ∠AOD = ∠BOC [Vertically opp. angles]

∠OAD = ∠OBC (Each = 90º) AD = BC (given) and By AAS congruence criterion, we have DAOD ≅ DBOC OA = OB (CPCT) i.e., O is the mid-point of AB.

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Hence, CD bisects AB. 4. Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore,

AD || BC and AB || CD. \ ABCD is a parallelogram.

i.e. AB = CD and BC = DA (Q opposite sides of a parallelogram are equal) Now, in Ds ABC and CDA, we have AB = CD (Proved above) BC = DA (Proved above) and AC = CA (common) \ By SSS criterion of congruence, we have DABC ≅ DCDA. 5. In Ds APB and AQB, we have ∠APB = ∠AQB (Q Each = 90º) ∠PAB = ∠QAB (Q AB bisects ∠PAQ) AB = AB (Common) By AAS congruence criterion, we have DAPB ≅ DAQB, which proves (i) BP = BQ (CPCT) i.e., B is equidistant from the arms of ∠A, which proves (ii) 6. In Ds ABC and ADE, we have AB = AD (given) ∠BAC = ∠DAE (Q ∠BAD = ∠EAC ⇒ ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠DAE) and AC = AE (given) \ By SAS criterion of congruence, we have DABC ≅ ∠ADE BC = DE (CPCT) 7. We have, ∠EPA = ∠DPB ∠EPA + ∠DPE = ∠DPB + ∠DPE ∠DPA = ∠EPB ....(1)

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Now, in Ds EBP and DAP, we have ∠EPB = ∠DPA [From (1)] BP = AP (Given) and ∠EBP = ∠DAP (Given) So, by ASA criterion of congruence, we have DEBP ≅ DDAP ⇒ BE = AD i.e. AD = BE (CPCT) 8. (i) In Ds AMC and BMD, we have AM = BM (Q M is the mid-point of AB) ∠AMC = ∠BMD (Vertically opp. angles) and CM = MD (given) \ By SAS criterion of congruence, we have \ DAMC ≅ DBMD (ii) Now, DAMC ≅ DBMD CA = DB ∠ACM = ∠BDM .... (1) (by CPCT) Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angles

∠BDM = ∠ACM are equal. Therefore, BD || CA. ∠CBD + ∠BCA = 180º (Q sum of the interior angles on the same side of transversal = 180º) ∠CBD + 90º = 180º (Q ∠BCA = 90º) ∠DBC = 90º (iii) Now, in Ds DBC and ACB, we have BD = CA [from (1)] ∠DBC = ∠ACB (Q Each = 90º) BC = CB (common) \ By SAS criterion of congruence, we have DDBC ≅ DACB CD = BA (CPCT) ½CD = ½AB CM = AB

Exercise 7.2 1. (i) In DABC, we have AB = AC ∠B = ∠C (Q Angles opposite to equal sides are equal)

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½∠B = ½∠C ∠OBC = ∠OCB … (i) (OB and OC bisect angles B and C respectively \ ∠OBC = ½∠B and ∠OCB = ½∠C) OB = OC (sides opp to equal angles are equal) … (ii)

(ii) Now, in Ds ABO and ACO, we have AB = AC (given) AO = AO (common) OB = OC [from (ii)] By SSS criterion of congruence, we have DABO ≅ DACO \ ∠BAO = ∠CAO (CPCT) \ AO bisects ∠BAC 2. In Ds ABD and ACD, we have DB = DC (given) ∠ADB = ∠ADC (Q AD ^ BC) AD = AD (common) \ By SAS criterion of congruence, we have DABD ≅ DACD AB = AC (CPCT) Hence, DABC is isosceles. 3. In DABC, AB = AC (Given) ∠ACB = ∠ABC .... (1) (Q Angles opposite to equal sides are equal) In DBFC & DCEB, we have ∠BFC = ∠CEB (Q Each = 90) ∠FBC = ∠ECB [from (1)] BC = CB (Common) \ By AAS criterion of congruence, we have DBFC ≅ DCEB

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\ CF = BE (Q Corresponding parts of congruent triangles are equal) 4. Let BE ^ AC and CF ^ AB. In Ds ABE and ACF, we have ∠AEB = ∠AFC (Q Each = 90º) ∠A = ∠A (common) and BE = CF (given) \ By AAS criterion of congruence, we have, DABE ≅ DACF AB = AC (Q Corresponding parts of congruent triangles are equal) 5. In DABC, we have AB = AC ∠ABC = ∠ACB ... (1) (Q Angles opposite to equal sides are equal) In DBCD, we have BD = CD ∠DBC = ∠DCB ... (2) (Q Angles opposite to equal sides are equal)

Adding (1) and (2), we have ∠ABC + ∠DBC = ∠ACB + ∠DCB ∠ABD = ∠ACD Alternate Method: In DABD & DACD,

AB = AC (Given) BD = DC (Given) AD = AD (common) DABD ≅ DACD (SSS criterion)

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\ ∠ABD = ∠ACD (cpct) 6. In DABC, we have AB = AC ∠ACB = ∠ABC ... (1) (Q Angles opposite to equal sides are equal) Now, AB = AD (given) \ AD = AC (Q AB = AC) Thus, in DADC, we have AD = AC ∠ACD = ∠ADC ...(2) (Q Angles opposite to equal sides are equal) Adding (1) and (2), we get ∠ACB + ∠ACD = ∠ABC + ∠ADC ∠BCD = ∠DBC + ∠BDC (Q ∠ADC = ∠BDC) (∠ABC = ∠DBC) ∠BCD + ∠BCD = ∠DBC + ∠BDC + ∠BCD (Adding ∠BCD on both sides) 2∠BCD = 180º (Angle sum property) ∠BCD = 90º Hence, ∠BCD is a right angle. 7. We have ∠A = 90º AB = AC ∠B = ∠C (Q Angles opposite to equal sides are equal)

Also, ∠A + ∠B + ∠C = 180º (Angle sum property) ⇒ 90º + 2∠B = 180º (Q ∠C = ∠B) ⇒ 2∠B = 180º – 90º = 90º ⇒ ∠B = 45º \ ∠C = ∠B = 45º 8. Let DABC be an equilateral triangle so that AB = AC = BC. Now, AB = AC \ ∠B = ∠C ... (1) (Q Angles opp. to equal sides are equal) Also Q CB = CA \ ∠A = ∠B ... (2) (Q Angles opp. to equal sides are equal)

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From (1) and (2), we have ∠A = ∠B = ∠C Also ∠A + ∠B + ∠C = 180º (Angle - sum property) ∠A + ∠A + ∠A = 180º 3∠A = 180º ∠A = 60º ∠A = ∠B = ∠C = 60º Thus, each angle of an equilateral triangle is 60º

Test Yourself – TR 2 2. a = 36º, b = 68º, c = 44º.

Exercise 7.3 1. (i) In Ds ABD and ACD, we have AB = AC (given) BD = DC (given) and AD = AD (common) \ By SSS criterion of congruence, we have DABD ≅ DACD ∠BAD = ∠CAD (CPCT) (ii) In Ds ABP and ACP, we have AB = AC ∠BAP = ∠PAC AP = AP (common) \ By SAS criterion of congruence, we have DABP ≅ DACP (iii) Since ∠BAD = ∠DAC AD bisects ∠A ⇒ AP bisects ∠A ... (1) In Ds BDP and CDP, we have BD = CD (given) BP = CP (Q DABP ≅ DACP ⇒ BP = CP)

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and DP = DP (common) \ By SSS criterion of congruence, we have DBDP ≅ DCDP ⇒ ∠BDP = ∠CDP (CPCT) ⇒ DP bisects ∠D ⇒ AP bisects ∠D ... (2) combining (1) and (2), we get AP bisects ∠A as well as ∠D. (iv) Since AP stands on BC ∠APB + ∠APC = 180º (linear pair) ∠APB = ∠APC (proved above)

∠APB = ∠APC = 1802

= 90º

BP = PC (proved above) ⇒ AP is the perpendicular bisector of BC. 2. AD is the altitude drawn from vertex A of an isosceles DABC to the opposite base BC so that

AB = AC, ∠ADC = ∠ADB = 90º. Now, in DADB and DADC, we have ∠ADC = ∠ADB (Q Each = 90º) AD = AD (common) and Hyp. AB = Hyp. AC (given) \ By RHS criterion of congruence, we have DADB ≅ DADC BD = CD ∠BAD = ∠CAD (CPCT) Hence, AD bisects BC, which proves (i), and AD bisects ∠A, which proves (ii) 3. Two Ds ABC and PQR in which AB = PQ, BC = QR and AM = PN. Since AM and PN are

medians of Ds ABC and PQR respectively, \ BM = QN ...(1)

Now, in Ds ABM and PQN, we have AB = PQ (given) BM = QN [from (1)]

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and AM = PN (given) \ By SSS criterion of congruence, we have DABM ≅ DPQN, which proves (i) ∠B = ∠Q ... (2) (CPCT) Now, in Ds ABC and PQR, we have AB = PQ (given) ∠B = ∠Q [from (2)] BC = QR (given) \ By SAS criterion of congruence, we have DABC ≅ DPQR, which proves (ii) 4. In Ds BCF and CBE, we have ∠BFC = ∠CEB (Q Each = 90º) Hyp. BC = Hyp. BC (common) FC = EB (given) \ By R.H.S criterion of congruence, we have DBCF ≅ DCBE \ ∠B = ∠C .... (1) (CPCT) Now, in DABC ∠B = ∠C [from (1)] \ AB = AC (Q sides opposite to equal angles of a triangle are equal) \ DABC is an isosceles triangle. 5. In Ds ABP and ACP, we have ∠APB = ∠APC (Q Each = 90º) AB = AC (given) and AP = AP (common)

\ By R.H.S criterion of congruence, we have DABP ≅ DACP ∠B = ∠C (CPCT)

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Test Yourself – TR 3 2. 1 : 1

Exercise 7.4 1. Let ABC be a right angled D in which ∠ABC = 90º. But ∠ABC + ∠BCA + ∠CAB = 180º 90º + ∠BCA + ∠CAB = 180º ∠BCA + ∠CAB = 90º ∠BCA and ∠CAB are acute angles ∠BCA < 90º and ∠CAB < ∠ABC AC > AB and AC > BC (Q side opposite to greater angle is larger) \In a right triangle, the hypotenuse is the longest side. 2. ∠PBC + ∠ABC = 180º (Linear pair) ⇒ ∠PBC = 180° – ∠ABC … (i) ∠QCB + ∠ACB = 180° (Linear pair) ⇒ ∠QCB = 180° – ∠ACB … (ii) As ∠PBC < ∠QCB ⇒ 180° – ∠ABC < 180° – ∠ACB ⇒ 180° + ∠ACB < 180° + ∠ABC ⇒ ∠ACB < ∠ABC ⇒ AB < AC (Q side opposite to smaller angle is smaller) 3. Since ∠B < ∠A and ∠C < ∠D \ AO < BO and OD < OC Adding these results, we have AO + OD < BO + OC AD < BC 4. ABCD is a quadrilateral such that AB is its smallest side and CD is its largest side. Join AC and BD. Since AB is the smallest side of quadrilateral ABCD.

\ In DABC, we have

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BC > AB ∠8 > ∠3 ... (1) (Q Angle opp. to longer side is greater) Since CD is the longest side of quadrilateral ABCD. In DACD, we have CD > AD ∠7 > ∠4 ... (2) (Q Angle opp. to longer side is greater) Adding (1) and (2), we get ∠8 + ∠7 > ∠3 + ∠4 ∠A > ∠C Again, in DABD, we have AD > AB (AB is the longest side) ∠1 > ∠6 ... (3) In DBCD, we have CD > BC (CD is the longest side) ∠2 > ∠5 ... (4) Adding (3) and (4), we get ∠1 + ∠2 > ∠6 + ∠5 ∠B > ∠D Thus, ∠A > ∠C and ∠B > ∠D 5.

(i) PR > PQ (ii) ∠PQR > ∠PRQ (angle opposite to larger side is greater (iii) ∠QPS = ∠RPS (given) (iv) ∠PQS + ∠QPS + ∠PSQ = 180° (angle sum property) (v) ∠PSR + ∠PRS + ∠SPR = 180° (angle sum property) ∠PQS = 180° – [∠QPS + ∠PSQ] ∠PRS = 180° – [∠PSR + ∠SPR] As PR > PQ 180° – [∠QPS + ∠PSQ] > 180° – [∠PSR + ∠SPR] 180° + (∠PSR + ∠SPR) > 180° + (∠QPS + ∠PSQ) (Q ∠QPS = ∠SPR)

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∠PSR > ∠PSQ 6. Let P be any point not on the straight line l. PM ^ l and N is any point on l other than M.

In DPMN, we have ∠M = 90º ∠N < 90º (∠M = 90º ⇒ ∠MPN + ∠PNM = 90º) ⇒ ∠P + ∠N = 90º ⇒ ∠N < 90º ∠N < ∠M PM < PN (Q Side opp. to greater angle is larger) Hence, PM is the shortest of all line segments from P to AB.

Test Yourself – TR 4 1. AC is longest and BC is shortest. 2. 1 : 1

Exercise 7.5 (Optional) 1. Let OD and OE be the perpendicular bisector of sides BC and CA of DABC.

\ O is equidistant from two ends B and C of line segment BC as O lies on the perpendicular bisector of BC. Similarly, O is equidistant from C and A.

Thus, the point of intersection O of the perpendicular bisector of sides BC, CA is the required point.

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2. Let BE and CF be the bisectors of ∠ABC and ∠ACB respectively intersecting AC and AB at E and F respectively.

Since O lies on BE, the bisector of ∠ABC, hence O will be equidistant from AB and BC. Again, O lies on the bisector CF of ∠ACB. Hence, O will be equidistant form BC and AC. Thus, O will be equidistant from AB, BC and CA.

3. The parlour should be equidistant from A, B and C, for which the point of intersection of perpendicular bisector should be located.

Thus, O is the required point which is equidistant from A, B and C. 4.

(i) Hexagon has 6 triangles and each triangle as shown above has 25 equilateral triangles of side 1 cm each. Hence 150 such triangles will be filled.

(ii) Star shaped Rangoli has 12 triangles and each triangle as shown above has 25 equilateral triangles of side 1 cm each. Hence 300 such triangles will be filled.