In ∆ abc and ∆ rqp ab qr and ∠ b ∠ q the two triangles are congruent by sas axiom if_________

isosceles but not congruent

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congruent but not isosceles

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neither congruent nor isosceles

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Criteria for Congruence of Trianglesp

Axiom 1: Side-Angle-Side (SAS) congruence rule
(An axiom is a mathematical statement which is assumed to be true without proof.)

Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Given: Two triangles ABC and PQR such that AB = PQ, AC = PR and ∠BAC = ∠QPR.

To prove: ∆ ABC ≅ ∆ PQR.

Proof: This result cannot be proved with help of previously known results. So, this rule is accepted as an axiom.

Another way to check whether the given two triangles are congruent or not, we follow a practical approach.

Place ∆ABC over ∆PQR such that the side AB falls on side PQ, vertex A falls on vertex P and B on Q. Since ∠BAC = ∠QPR. Therefore, AC will fall on PR. But AC = PR and A 

falls on P. therefore, C will fall on R. Thus, AC coincides with PR.

Now, B falls on Q and C falls on Therefore, BC coincides with QR.

Thus, ∆ ABC when superposed on ∆ PQR, covers it exactly. Hence, by the definition of congruence, ∆ ABC ≅ ∆ PQR.

Example 1: Check whether ∆ABC and ∆PQR are congruent or not.

Solution: In ∆ ABC and ∆ PQR, we have

   AB = PQ = 5 cm (Given)

∠ BAC = ∠QPR = 40o (Given)

AC = PR = 4 cm (Given)

Therefore, ∆ ABC ≅ ∆ PQR (By SAS-criterion of congruence)

Example 2: In the figure below, R is the mid-point of PT and SQ. Prove that ∆ PQR ≅ ∆ TSR.

Given: PR = RT and SR = RQ.

To prove: ∆ PQR ≅ ∆ TSR.

Proof: In ∆ PQR and ∆ TSR, we have

PR = TR                               (R is the mid-point of PT)

∠PRQ = ∠TRS                     (Vertically opposite angles are equal)

QR = SR                               (R is the mid-point of SQ)

Therefore, Δ PQR ≅ Δ TSR (By SAS-criterion of congruence)

Example 3: In the figure, it is given that PT = PU and QT = RU. Prove that ΔPTR ≅ ΔPUQ

Given: PT = PU and QT = RU.

To prove: Δ PTR ≅ Δ PUQ.

Proof: We have,

                              PT = PU                           ........... (I)

 And,                      QT = RU                          .......... (II)

Adding equation (I) and (II), We get,

                                 PT + QT = PU + RU

⇒                               PQ = PR                     ........... (III)

Now, in ∆ PTR and ∆ PUQ, we have

                                   PT = PU [Given]

⇒                             ∠ TPR = ∠UPQ                     [Common]

⇒                             PQ = PR                                [From (III)]

Therefore, Δ PTR ≅ Δ PUQ [By SAS-criterion of congruence]

Theorem 1: Angle-Side-Angle (ASA) Congruence rule Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the 

included side of the other triangle.

Given: ΔPQR and ΔMNO such that ∠PQR = ∠MNO, ∠PRQ = ∠MON and QR = NO.

To prove: Δ PQR ≅ Δ MNO.

Proof: There are three possibilities that arise.

CASE I: When PQ = MN

In this case, we have

                               PQ = MN

                             ∠PQR = ∠MNO            (Given)

                               QR = NO                    (Given)

Therefore, Δ PQR ≅ Δ MNO (By SAS-criterion of congruence)

CASE II: PQ < MN

Construction: Join OS such that NS = PQ.

In Δ PQR and Δ SNO, we have

                                PQ = SN

                                 ∠PQR = ∠MNO      (Given)

                                      QR = NO            (Given)

Therefore, Δ PQR ≅ Δ SNO (By SAS-criterion of congruence)

By using corresponding parts of congruent triangles

⇒                             ∠PRQ = ∠SON.

But,                          ∠PRQ = ∠MON. (Given)

This is possible only when ray SO coincides with ray MO or S coincides with M. Therefore, PQ must be equal to MN.

∴                              ∠SON = ∠MON.

Thus, in Δ PQR and Δ MNO, we have

                                    PQ = MN

⇒                              ∠PQR = ∠MNO (Given)

⇒                                QR = NO (Given)

Therefore, Δ PQR ≅ Δ MNO (By SAS-criterion of congruence)

CASE III: PQ > MN.

Construction: Join SO such that NS = PQ.

In Δ PQR and Δ SNO, we have

                                  PQ = SN

                               ∠PQR = ∠MNO (Given)

                                 QR = NO (Given)

Therefore, Δ PQR ≅ Δ SNO (By SAS-criterion of congruence)

By using corresponding parts of congruent triangles

⇒                            ∠PRQ = ∠SON.

But,                        ∠PRQ = ∠MON. (Given)

This is possible only when ray SO coincides with ray MO or S coincides with M. Therefore, PQ must be equal to MN.

∴                             ∠SON = ∠MON.

Thus, in Δ PQR and Δ MNO, we have

PQ = MN

⇒ ∠PQR = ∠MNO (Given)

⇒ QR = NO (Given)

Therefore, Δ PQR ≅ Δ MNO (By SAS-criterion of congruence

Hence, in all the three cases, we have Δ PQR ≅ Δ MNO.

Example 1: Check whether ∆ABC ≅ ∆PQR?

Solution: In ∆ABC and ∆PQR, we have

                            ∠ABC = ∠PQR = 40o          (Given)

                               BC = QR = 5 cm            (Given)

                            ∠BCA = ∠QRP = 60o          (Given)

Therefore, ∆ ABC ≅ ∆ PQR (By ASA-criterion of congruence)

Example 2: In the figure, ST ∥ QP and R is the mid-point of SQ, prove that R is also the mid-point of PT.

Given: ST ∥ QP and R is the mid-point of SQ.

To prove: PR = RT.

Proof: Since QP ∥ ST and transversal PT cuts them at P and T respectively.

∴                                  ∠RTS = ∠RPQ.                              (Alternate interior angles) ......... (I)

Similarly, ∠RST = ∠RQP.                                                 (Alternate interior angles) ......... (II)

Since PT and SQ intersect at R.

∴                                 ∠QRP = ∠SRT.                            (Vertically opposite angles) ...........(III)

Thus, in Δ PQR and Δ SRT, we have

                                   ∠ PQR = ∠ RST [From (II)]

⇒                                  QR = RS [Given]

⇒                                ∠ QRP = ∠ SRT [From (III)]

Therefore, Δ PQR ≅ Δ SRT [By ASA-criterion of congruence]

By using corresponding parts of congruent triangles

⇒                                  PR = RT.

Hence, R is the mid-point of PT.

Theorem 2: Angle-Angle-Side (AAS) Congruence rule If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

Given: In ΔPQR and ΔMNO, ∠PQR = ∠MNO, ∠QPR = ∠NMO & QR=NO.

To prove: ΔPQR ≅ ΔMNO

Proof: We have,

                                   ∠PQR = ∠MNO and ∠QPR = ∠NMO

⇒                          ∠PQR + ∠QPR = ∠NMO + ∠MNO                          .................. (I)

∵                          ∠PRQ + ∠PQR + ∠ QPR = 180°   [Sum of angles of a triangle]

∴                         ∠ PQR + ∠ QPR = 180° − ∠PRQ                             ................. (II)

Similarly, ∠ NMO + ∠ MNO = 180° − ∠MON                                     ................. (III)

From equation (I), (II) & (III), we have

                               180° − ∠PRQ = 180° − ∠MON

⇒                           ∠PRQ = ∠MON                                                     ................ (IV)

Now, in ΔPQR and ΔMNO, we have

                             ∠PQR = ∠MNO                             [Given]

                              QR = NO                                      [Given]

                             ∠PRQ = ∠MON                         [From (IV)]

Therefore, Δ PQR ≅ Δ MNO [By ASA-criterion of congruence]

Example1: Check whether two triangles ABC and PQR are congruent.

Solution: In ∆ ABC and ∆ PQR, we have

                             ∠ABC = ∠PQR = 40o                    (Given)

                            ∠BCA = ∠QRP = 60o                     (Given)

                              AC = PR = 3 cm                           (Given)

Therefore, ∆ ABC ≅ ∆ PQR (By AAS-criterion of congruence)

Example 2: PQ and RS are perpendiculars of equal length, to a line segment PS.

Show that QR bisects PS. 

Given: PQ = RS, PQ ⊥ PS and RS ⊥ PS.

To prove: OP = OS.

Proof: In triangles OPQ and ORS, we have

                                ∠POQ = ∠SOR                            [Vertically opposite angles]

                                ∠OPQ = ∠OSR                             [Each equal to 90°]

                                  PQ = RS                                           [Given]

                           Therefore, Δ POQ ≅ ΔSOR              [By AAS-criterion of congruence]

By using corresponding parts of congruent triangles

⇒                                  OP = OS.

Hence, O is the mid-point of PS.

Theorem 3: Side-Side-Side (SSS) Congruence rule If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Given that: Two Δ ABC and Δ DEF such that AB = DE, BC = EF and AC = DF.

To prove: Δ ABC ≅ Δ DEF.

Construction: Suppose BC is the longest side. Draw EG such that

∠ FEG = ∠ ABC and EG = AB. Join GF and GD

Proof: In ΔABC and ΔGEF, we have

                                   BC = EF                               [Given]

                                   AB = GE                               [By Construction]

                                   ∠ABC = ∠GEF                    [By Construction]

Therefore, Δ ABC ≅ Δ GEF [By SAS-criterion of congruence]

By using corresponding parts of congruent triangles

⇒                              ∠BAC = ∠EGF and AC = GF

Now, AB = DE and AB = GE

⇒                                  DE = GE                                    .................. (I)

Similarly, AC = DF and AC = GF

⇒                                   DF = GF                                   .................. (II)

In Δ EGD, we have

                                      DE = GE                                     [From (I)]

Since angles opposite to equal sides of an isosceles triangle are equal.

⇒                                ∠EDG = ∠EGD                           ................ (III)

In Δ FGD, we have

                                   DF = GF                                       [From (II)]

Since angles opposite to equal sides of an isosceles triangle are equal.

⇒                                ∠FDG = ∠FGD                         ................ (IV)

From (III) and (IV), we have,

∠EDG + ∠FDG = ∠EGD + ∠FGD

⇒                                ∠EDF = ∠EGF

But,                           ∠BAC = ∠EDF                            ................. (V

In Δ ABC and Δ DEF, we have

                                   AC = DF                                          [Given]

⇒                               ∠BAC = ∠EDF                            [From (V)]

And,                            AB = DE                                      [Given]

Therefore, ΔABC ≅ ΔDEF [By SAS-criterion of congruence]

Example 1: Check whether two triangles ABC and PQR are congruent.

Solution: In ∆ ABC and ∆ PQR, we have                                 BC = QR                (Given) ⇒                             AB = PQ                 (Given)

⇒                             AC = PR                  (Given)

Therefore, ∆ ABC ≅ ∆ PQR (By SSS-criterion of congruence)

Example 2: In the figure, it is given that PR = QS and PS = RQ. Prove that Δ SPR ≅ Δ RQS.

Proof : In triangles SPR and RQS, we have

                                PR = QS                     [Given] ⇒                            PS = QR                     [Given]

⇒                            RS = SR                      [Given]

Therefore, Δ SPR ≅ ΔRQS [By SSS-criterion of congruence]

Theorem 4: Right angle -Hypotenuse-Side (RHS) Congruence rule If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Given that: Two right triangles PQR and SUT in which ∠PQR = ∠SU = 90°, PR = ST, QR = UT.

To prove: ΔPQR ≅ ΔSUT.

Construction: Produce SU to V so that UV = PQ. Join VT.

Proof: In Δ PQR and Δ SUT, we have

                              PQ = VU                          [By construction]

                            ∠PQR = ∠VUT                 [Each equal to 90°]

                             QR = UT                            [Given]

Therefore, ΔPQR ≅ ΔSUT [By SAS-criterion of congruence]

By using corresponding parts of congruent triangles

⇒                            ∠QPR = ∠UVT                      .............. (I)

And,                           PR = VT

⇒                               PR = ST                              ............ (II)

∴                                 VT = ST

Since, angles opposite to equal sides in Δ STV are equal.

⇒                               ∠UST = ∠TVS                   ............. (III)

From (I) and (III), we get,

                               ∠QPR = ∠UST                     .............. (IV) 

And given that,     ∠RQP = ∠TUS                       .............. (V)

Adding (IV) & (V), we get,

                           ∠QPR + ∠RQP = ∠UST + ∠TUS ............. (VI)

∵ ∠PRQ + ∠RQP + ∠QPR = 180°

∴                             ∠QPR + ∠RQP = 180 - ∠PRQ ........... (VII)

Similarly,               ∠UST + ∠TUS = 180° - ∠STU ............ (VIII)

From equation (VI), (VII) & (VIII), we have

                              180° - ∠PRQ = 180° - ∠STU

⇒                           ∠PRQ= ∠STU.                       ............. (IX)

Now, in Δ PQR and Δ SUT

                               QR = UT                               [Given]

⇒                          ∠PRQ = ∠STU.                  [From (IX)]

And,                        PR = ST                             [Given]

Therefore, ΔPQR ≅ ΔSUT [By SAS-criterion of congruence]

Example 1: Check whether two triangles ABC and PQR are congruent.

Solution: In ∆ ABC and ∆ PQR, we have 

                  ∠ ABC = ∠ PQR = 90o                           (Given)

                      AC = PR                                             (Given)
                      AB = PQ                                             (Given)

Therefore, ∆ ABC ≅ ∆ PQR (By RHS-criterion of congruence)

Example 2: In the figure below, it is given that LM = MN, BM = MC, ML ⊥ AB and MN ⊥ AC. Prove that AB = AC.

Proof: In right-angled Δ BLM and Δ CNM, we have 

                         BM = MC                                       [Given]

⇒                     LM = MN                                        [Given]

Therefore, Δ BLM ≅ Δ CNM [By RHS-criterion of congruence]

By using corresponding parts of congruent triangles

⇒                      ∠LBM = ∠NCM
Since, sides opposite to equal angles are equal.

Therefore, AB = AC.

Some more rules

     1. Angle-Angle-Angle (AAA) Rule

In Δ ABC and Δ PQR, we have

                         ∠BAC = ∠QPR                (Given)                         ∠ACB = ∠PRQ                 (Given)

                        ∠CBA = ∠RQP                  (Given)

But, Δ ABC and Δ PQR are similar but not congruent because their sizes are different.

2. Angle-Side-Side (ASS or SSA) Rule

We have a triangle ABD. We draw AC such that AC = AD.

Now, consider the two triangles, Δ ABD and Δ ABC

Now, let us check whether we can use ASS criteria for the

congruency of two triangles or not. In Δ ABD and Δ ABC, we have

                                ∠ABD = ∠ABC                                (Given)                                   AB = AB                                        (Common)

                                   AD = AC                                         (Given)

So, the corresponding Angle-Side-Side of the two triangles are equal but, these two figures are different in shape.

So, we can conclude that this method is not a universal method for proving triangles congruent.

Page 2

Some Properties of a Triangle

Theorem 5: Angles opposite to equal sides of an isosceles triangle are equal.

Given: ∆PQR is an isosceles triangle in which PQ = PR.

To prove: ∠ PQS = ∠PRS.

Construction: Draw the bisector PS of ∠QPR which meets QR in S.

Proof: In ∆ PQS and ∆ PRS, we have

                          PQ = PR                                     (Given)

⇒                  ∠ QPS = ∠ RPS                              (By construction)

⇒                       PS = PS                                      (Common)

Therefore, ΔPQS ≅ ΔPRS (By SAS-criterion of congruence)

By using corresponding parts of congruent triangles

⇒                     ∠PQS = ∠PRS.

Theorem 6: The sides opposite to equal angles of a triangle are equal.

Given: In triangle PQR, ∠PQR = ∠PRQ.

To prove: PQ = PR.

Construction: Draw the bisector of ∠QPR and let it meet QR at S.

Proof: In ∆ PQS and ∆ PRS, we have

                        ∠PQR = ∠PRQ                         (Given)

⇒                     ∠QPS = ∠RPS                         (By construction)

⇒                      PS = PS                                   (Common)

Therefore, ΔPQS ≅ ΔPRS ( By AAS-criterion of congruence)

By using corresponding parts of congruent triangles

⇒                      PQ = PR.

Example 1: In Δ PQR, ∠ QPR = 80° and PQ = PR. Find ∠RQP and ∠PRQ.

Given that: ∠QPR = 80° and PQ = PR.

Solution: We have

                               PQ = PR

Since angles opposite to equal sides are equal.

⇒                            ∠RQP = ∠PRQ

In Δ PQR, We have

         ∠QPR + ∠RQP + ∠PRQ = 180°.

⇒      ∠QPR + ∠RQP + ∠RQP = 180°. (∵ ∠ RQP = ∠PRQ)

⇒                      80° + 2 ∠RQP = 180°

⇒                               2 ∠RQP = 180° - 80°

⇒                                2 ∠RQP = 100°

⇒                                 ∠RQP =

⇒                                   ∠RQP = 50°

Hence, ∠RQP = ∠PRQ = 50°

Example 2: In figure, PQ = PR and ∠PRS = 110°. Find ∠P.

Solution: We have,

                                      PQ = PR

Since angles opposite to equal sides are equal.

⇒                              ∠PQR = ∠PRQ.

Now,             ∠PRQ + ∠PRS = 180° (∵ Linear pair of angles)

⇒                    ∠PRQ+ 110° = 180°

⇒                  ∠PRQ = 180° - 110°

⇒                   ∠PRQ = 70°

And also,      ∠PQR = 70°.

Now,    ∠QPR + ∠PQR + ∠PRQ = 180°

⇒                  ∠QPR+ 70° + 70° = 180° (∵ ∠PQR = ∠PRQ)

⇒                        ∠QPR + 140° = 180°

⇒                                    ∠QPR = 180°- 140°

⇒                                    ∠QPR = 40°.

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Inequalities in a Triangle

Theorem 7: If two sides of a triangle are unequal, the angle opposite to the longer side is larger.

Given that: In Δ MNO, MO > MN.

To prove: ∠ MNO > ∠ MON.

Construction: Mark a point P on MO such that MN = MP. Join NP.

Proof: In Δ MNP, we have

                              MN = MP                      [By construction]

Since angles opposite to equal sides are equal.

⇒                         ∠MNP = ∠MPN.                    ................ (I)

Now, consider Δ NPO. We find that ∠MPN is the exterior angle of Δ NPO and an exterior angle is always greater than interior opposite angles.

Therefore, ∠MPN > ∠PON

⇒                     ∠MPN > ∠MON                     ................ (II)

From (I) and (II), we have

                       ∠MNP > ∠MON                     ................ (III)

But, ∠MPN is a part of ∠ MNO.

∴                    ∠ MNO > ∠MNP                    ................. (IV)

From (III) and (IV), we get

                     ∠ MNO > ∠MON.

Theorem 8: In any triangle, the side opposite to the larger (greater) angle is longer.

Given that: In In Δ MNO, ∠ MNO > ∠MON.

To prove: MO > MN.

Proof: In Δ MNO, we have the following three possibilities.

                  (I) MO = MN

                  (II) MO < MN

                  (III) MO > MN

Out of these three possibilities exactly one must be true.

CASE I: When MO = MN

Since angles opposite to equal sides are equal.

⇒                  ∠MNO = ∠MON

This is a contradiction that ∠ MNO > ∠MON.

∴ MO ≠ MN.

CASE II: When MO < MN

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

⇒ ∠MNO < ∠MON

This also contradicts the given hypothesis that ∠MNO > ∠MON

Thus, we are left with the only possibilities, MO > MN, which must be true.

Hence, MO > MN.

Example: In a Δ MNO, if ∠MON = 55° and ∠ONM = 66°, determine

the shortest and largest sides of the triangle.

Solution: We have,

∠MON= 55° and ∠ONM = 66°

∴ ∠MON + ∠ONM + ∠NMO = 180°

⇒          55° + 66° + ∠NMO = 180°

⇒                  121° + ∠NMO = 180°

⇒                              ∠NMO = 180° - 121°

⇒                              ∠NMO = 59°.

Since ∠NMO and ∠ONM are the smallest and largest angles respectively. Therefore, sides ON and OM are the smallest and largest sides respectively of the triangle.

Theorem 9: The sum of any two sides of a triangle is greater than the third side.

Given: MNO is a triangle.

To prove: (I) MN + MO > NO

                  (II) MN + NO > MO

                  (III) NO + MO > MN

Proof: (I) Produce side NM to P such that MP = MO. Join PO

           In Δ MOP, we have

                  MP = MO     (By construction)

Since angles opposite to equal sides are equal.

⇒             ∠MOP = ∠MPO

⇒    ∠NOM + ∠MOP > ∠MPO

⇒                   ∠NOP > ∠MPO

Since side opposite to greater angle is larger.

∴                     NP > NO

⇒          MN + MP > NO

⇒          MN + MO > NO         [∵ MO = MP by construction]

Thus,    MN + MO > NO.

(II) Produce side MN to P such that NP = NO. Join PO.

     In Δ NOP, we have

                      NP = NO              (By construction)

Since angles opposite to equal sides are equal.

⇒                   ∠NOP = ∠NPO

⇒     ∠NOM + ∠NOP > ∠NPO

⇒                    ∠MOP > ∠NPO

Since side opposite to greater angle is larger. ∴                             MP > MO ⇒                   MN + NP > MO ⇒                   MN + NO > MO              [∵ NO = NP by construction]

Thus,             MN + NO > MO.

(III) Produce side MO to P such that OP = NO. Join NP

                 In Δ NOP, we have
                      OP = ON            (By construction)

Since angles opposite to equal sides are equal.

⇒                        ∠ONP = ∠OPN ⇒          ∠ONM + ∠ONP > ∠OPN

⇒                         ∠MNP > ∠OPN

Since, side opposite to greater angle is larger.

∴                                       MP > MN ⇒                             MO + OP > MN ⇒                             MO + NO > MN       [∵ NO = OP by construction]

Thus,                       MO + NO > MN.

Hence, MN + MO > NO , MN + NO > MO and NO + MO > MN.

Example: In Δ MNO, P is any point on the side NO. Show that MN + MO + NO > 2 MP.

Solution: In Δ MNP, we have

                            MN + NP > MP                     ............... (I)

(∵ Sum of two sides of a triangle is greater than the third side)

Similarly, In Δ MOP, we have

                           MO + OP > MP                    ............... (II)

Adding (I) and (II), we get

     (MN + NP) + (MO + OP) > MP + MP

⇒ MN + (NP + OP) + MO > 2 MP

⇒             MN + NO + MO > 2 MP.

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