Definitions [3]
SAS Congruence criterion: If under a correspondence, two sides and the angle included between them of a triangle are equal to two corresponding sides and the angle included between them of another triangle, then the triangles are congruent.
ASA Congruence criterion: If under a correspondence, two angles and the included side of a triangle are equal to two corresponding angles and the included side of another triangle, then the triangles are congruent.
RHS Congruence criterion: If under a correspondence, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the triangles are congruent.
Theorems and Laws [2]
In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥ PQ;
Prove that:
- ΔXTQ ≅ ΔXSQ.
- PX bisects angle P.
Given: A(ΔPQR) in which QX is the bisector of ∠Q and RX is the bisector of ∠R.
XS ⊥ QR and XT ⊥ PQ.
We need to prove that:
- ΔXTQ ≅ ΔXSQ.
- PX bisects angle P.
Construction: Draw XZ ⊥ PR and join PX.
i. In ΔXTQ and ΔXSQ,
∠QTX = ∠QSX = 90° ...[XS ⊥ QR and XT ⊥ PQ]
∠TQX = ∠SQX ...[QX is bisector of ∠Q]
QX = QX ...[Common]
∴ By Angle-Side-Angle Criterion of congruence,
ΔXTQ ≅ ΔXSQ
ii. The corresponding parts of the congruent triangles are congruent.
∴ XT = XS ...[c.p.c.t.]
In ΔXSR and ΔXRZ
∠XSR = ∠XZR = 90° ...[XS ⊥ QR and ∠XSR = 90°]
∠XRS = ∠ZRX ...[RX is bisector of ∠R]
RX = RX ....[Common]
∴ By Angle-Angle-Side criterion of congruence,
ΔXSR ≅ ΔXRZ
The corresponding parts of the congruent triangles are congruent.
∴ XS = XT ...[c.p.c.t.]
From (1) and (2)
XT = XZ
In ΔXTP and ΔPZX
∠XTP = ∠XZP = 90° ....[Given]
XP = XP ....[Common]
XT = XZ
∴ By Right angle-Hypotenuse-side criterion of congruence,
ΔXTP ≅ ΔPZX
The corresponding parts of the congruent triangles are
congruent.
∴ ∠TPX = ∠ZPX ...[c.p.c.t.]
∴ PX bisects ∠P.
Theorem (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 other triangle.
Proof : We are given two triangles ABC and DEF in which:
∠ B = ∠ E, ∠ C = ∠ F
and BC = EF
We need to prove that ∆ ABC ≅ ∆ DEF
For proving the congruence of the two triangles see that three cases arise.
Case (i) : Let AB = DE in following fig.
You may observe that
AB = DE (Assumed)
∠ B = ∠ E (Given)
BC = EF (Given)
So, ∆ ABC ≅ ∆ DEF (By SAS rule)
Case (ii) :
Let if possible AB > DE. So, we can take a point P on AB such that PB = DE. Now consider ∆ PBC and ∆ DEF in following fig.
Observe that in ∆ PBC and ∆ DEF,
PB = DE (By construction)
∠ B = ∠ E (Given)
BC = EF (Given)
So, we can conclude that:
∆ PBC ≅ ∆ DEF, by the SAS axiom for congruence.
Since the triangles are congruent, their corresponding parts will be equal.
So, ∠ PCB = ∠ DFE
But, we are given that
∠ ACB = ∠ DFE So, ∠ ACB = ∠ PCB
This is possible only if P coincides with A.
or, BA = ED
So, ∆ ABC ≅ ∆ DEF (by SAS axiom)
Case (iii) : If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in Case (ii), we can conclude that AB = DE and so, ∆ ABC ≅ ∆ DEF.
You know that the sum of the three angles of a triangle is 180°. So if two pairs of angles are equal, the third pair is also equal (180° – sum of equal angles).
So, two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. We may call it as the AAS Congruence Rule.
Key Points
-
AA / AAA → two angles equal
-
SAS → included angle equal + sides proportional
-
SSS → all sides proportional
