CONGRUENCE OF TRIANGLES - हिन्दी ... · 9/10/2009 · AD is an altitude of an isosceles...

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CONGRUENCE OF TRIANGLES

1. Two figures are congruent, if they are of the same shape and of the same size.2. Two circles of the same radii are congruent.3. Two squares of the same sides are congruent.4. If two triangles ABC and PQR are congruent under the correspondence A – P,B-Q and C-R, then symbolically, it is expressed as Δ ABC Δ PQR.

5. SAS Congruence Rule: If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent. (Axiom: This result cannot be proved with the help of previously known results.)

6. ASA Congruence Rule: If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent (ASA Congruence Rule).

Construction: Two triangles are given as follows, where DEFABC and DFEACB . Sides AB=DE

To Prove: DEFABC

Proof: DEFABC (given)AB = DEAC= DF (Sides opposite to corresponding angles are in the same ratio as ratio

of angles)

Hence, by SAS congruence rule DEFABC is proved.

A

B C



D

E F

7. AAS Congruence Rule: If two angles and one side of one triangle are equal totwo angles and the corresponding side of the other triangle, then the two triangles are congruent.

This theorem can be proved in similar way as the previous one.

8. Angles opposite to equal sides of a triangle are equal.9. Sides opposite to equal angles of a triangle are equal.10. Each angle of an equilateral triangle is of 60°.

11. SSS Congruence Rule: If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent.

12. RHS Congruence Rule: If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS Congruence Rule).

13. In a triangle, angle opposite to the longer side is larger (greater).14. In a triangle, side opposite to the larger (greater) angle is longer.15. Sum of any two sides of a triangle is greater than the third side.

Theorem: Angles opposite to equal sides of an isosceles triangle are equal.Theorem: The sides opposite to equal angles of a triangle are equal.Theorem: If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).Theorem: In any triangle, the side opposite to the larger (greater) angle is longer.Theorem: The sum of any two sides of a triangle is greater than the third side.



EXERCISE 1

1. In quadrilateral ACBD, AC = AD and AB bisects A. Show that Δ ABC Δ ABD.

Answer: In ADBACB &AC=AD

DABCAB (AB is bisecting CAD )AB = AB (common side in both triangles)

So, by SAS axiom it is proved that;ADBACB

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

Answer:

(i) In BACABD &AD=BCAB=AB (common)

ABCBAD

So, by SAS rule BACABD

(ii) Since, BACABD , so BD = AC (third corresponding sides of respective triangles).

(iii) In congruent triangles all corresponding angles

Are always equal, so ABCBAD is proved

3. AD and BC are equal perpendiculars to a linesegment AB. Show that CD bisects AB.

Answer: In AODBOC &BC=AD (given)

DAOCBO (Right Angle)AODBOC (Opposite angles of intersecting lines

So, by ASA rule AODBOC AOBO and it is proved that CD bisects AB.

A B

C

D

A

B

C

D

A

BC

D

O



4. l and m are two parallel lines intersected byanother pair of parallel lines p and q. Show that Δ ABC Δ CDA.

Answer: In CDAABC &AB = CD (l and m are parallel)AD=BC (AB and CD are parallel)

DCmABC (Angles on the same side of transversal BC)

ADCDCm (Alternate Angles are equal)

So, ADCABC So, by SAS rule CDAABC

5. Line l is the bisector of an angle A and B is anypoint on l. BP and BQ are perpendiculars from Bto the arms of A . Show that:(i) Δ APB Δ AQB(ii) BP = BQ or B is equidistant from the armsof A .

Answer: In AQBAPB &AB=AB (Common side)

QABPAB (AB is bisector of QAP )APBAQB (Right Angle)

So, by ASA rule AQBAPB And BQ=BP

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

Answer: In ADEABC &AB=AD (given)AC=AE (given)

Since, EACBAD So, DACEACDACBAD Or, DAEBAC So, by SAS rule ADEABC

DEBC proved

7. AB is a line segment and P is its mid-point. D andE are points on the same side of AB such that BAD = ABE and EPA = DPB

A

B C

D l

p q

m

AB

P

Q l

A

B CD

E

A BP

DE



Show that(i) Δ DAP Δ EBP(ii) AD = BE

Answer: In EBPDAP ABEBAD (given)DPBEPA (given)

So, EPDDPBEPDEPA Or, EPBDPA AP=PB (Since P is mid point)So, by ASA rule EBPDAP So, AD=BE

8. In right triangle ABC, right angled at C, M isthe mid-point of hypotenuse AB. C is joinedto M and produced to a point D such thatDM = 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 =2

1AB

Answer: In BMDAMC &BM=AM (M is midpoint)DM=CM (given)

AMCDMB (opposite angles)

So, BMDAMC Hence, DB=AC

BACDBA So, DB||AC (alternate angles are equal)

So, ACBBDC = Right Angle)(internal angles are complementary in Case of transversal of parallel lines)

ACBDBC &DB=AC (proved earlier)BC=BC (Common side)

ACBBDC (proved earlier)

So, ACBDBC So, AB=DCSo, AM=BM=CM=DM

A

B C

D

M



So, CM=2

1AB

EXERCISE 21. In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersecteach other at O. Join A to O. Show that :(i) OB = OC (ii) AO bisects A

Answer: In OBCOACOBC (they are half of angles B & C)

So, OB=OC ( Sides opposite to equal angles)

In AOCAOB &AB=AC (given)OB=OC (proved earlier)

ACOABO (they are half of angles B & C)

So, AOCAOB (SAS Rule)

So, CAOBAO It means that AO bisects A

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

Answer: In ACDABD &AD=AD (common side)BD=CD (given)

ADCADB (right angle)

So, ACDABD So, AB=AC, which proves that ABC is isosceles

3. ABC is an isosceles triangle in which altitudesBE and CF are drawn to equal sides AC and ABrespectively. Show that these altitudes are equal.

Answer: In ACFABE &AB=AC (given)

CAFBAE (common to both triangles)BEACFA (right angles)

A

B C

O

A

B CD

A

B C

F E



So, ACFABE ( ASA Rule ) So, BE=CF

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

Answer: This can be solved like previous question.

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

A

B C

D

Answer: ACBABC DCBDBC

So, DCBACBDBCABC ACDABD

6. ΔABC 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.

Answer: In ABCADC &AD=ABAC=AC

ABCACB ADCACD

In ΔABC, 180CABABCACB

A

D

BC



ACBCAB 2180 ------------------------------ (1)

Similarly in ΔADC, ACDDAC 2180 ------------(2)

As BD is a straight line, so 180DACCABSo, adding equations (1)&(2) we get

ACDACB 22360180)(2360180 ACDACB

180)(2 ACDACB 90BCDACDACB

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

Answer: If AB= AC then angles opposite to these sides will be equal. As you know the sum of all angles of a triangle is equal to 180°, So, A+ B+C=180°Or, 90°+ B+C=180°Or, B+C=180°-90°=90°Or, B=C=90°

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

Answer: As angles opposite to equal sides of a triangle are always equal. So, in case of equilateral triangle all angles will be equal. So they will measure one third of 180°, which is equal to 60°

EXERCISE 3

1. Δ ABC and Δ DBC are two isosceles triangles onthe same base BC and vertices A and D are on thesame side of BC . If AD is extendedto 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: In ACDABD &AB=ACBD=CDAD=ADSo, Δ ABD Δ ACD (SSS Rule)

In ACPABP &AB=ACAP=AP

B

A

CP

D



ACPABP (Angle opposite to equal sides)So, Δ ABP Δ ACP (SAS Rule)Since Δ ABP Δ ACP

So, CAPBAP So, AP is bisecting BACSimilarly CDPBDP & can be proved to be Congruent and as a result it can be proved that

AP is bisecting BDC

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: This can be solved like previous question.

3. Two sides AB and BC and median AMof one triangle ABC are respectivelyequal to sides PQ and QR and medianPN of Δ PQR. Show that:(i) Δ ABM Δ PQN(ii) Δ ABC Δ PQR

Answer: In PQNABM &AB=PQAM=PNBM=QN (median bisects the base)

So, PQNABM

In PQRABC &AB=PQBC=QRAC=PR (Equal medians means third side will be equal)

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

Answer: In AFCAEB &BE=CF (Perpendicular)AB=BC (Hypotenuse)

So, AFCAEB 5. ABC is an isosceles triangle with AB = AC. Draw AD BC to show that B = C.

Answer: After drawing AD BC

In ADBADC &AC=AB

B

A

C

P

MQ R

N

A

B C

F E

A

B CD



AD=ADADBADC

So, ADBADC So, ABCACD

EXERCISE 41. Show that in a right angled triangle, the hypotenuse is the longest side.

Answer: In a right angled triangle, the angle opposite To the hypotenuse is 90°, while other two angles are Always less than 90°. As you know that the side opposite to the largest angle is always the largest in a triangle.

2. In the given triangle sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.

Answer: PBCABC 180OCBACB 180

Since OCBPBC So, ACBABC As you know side opposite to the larger angle is largerthan the side opposite to the smaller angle.Hence, AC>AB

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

Answer: AO<BO (Side opposite to smaller angle)DO<CO (Side opposite to smaller angle)So, AO+DO<BO+COOr, AD<BC

4. AB and CD are respectively the smallest andlongest sides of a quadrilateral ABCD. Show that A > C and B > D.

Answer: Let us draw two diagonals BD and AC as shown in the figure.In ΔABD Sides AB<AD<BDSo, ABDADB ----------------- (1)Angle opposite to smaller side is smallerIn ΔBCD Sides BC<DC<BD

So, CBDBDC ---------------- (2)

B

A

C

P Q

A

B

C

D

O

A

B C

D



Adding equation (1) & (2)CBDABDBDCADB

ABCADC

Similarly in ABCACBBAC ------------------ (3)

In ADCDCADAC ----------------- (4)

Adding equations (3) & (4)DCAACBDACBAC

BCDBAD

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

P

Q RS

1 2

4 5

3

Answer: For convenience let us name these angles as follows:1PQR2PRQ3QPR4QPS5RPS6PSQ7PSR

Since, PR>PQ , so 21 In PQS



180641In PRS

180752

In both these triangles 54 21

So, for making the sum total equal to 180° the following will always be true:76

CONGRUENCE OF TRIANGLES - हिन्दी ... · 9/10/2009 · AD is an altitude of an isosceles...

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