Consider two given triangles PQR and ABC. They are said to be congruent (≡) if their corresponding parts are congruent. That is PQ=AB, QR=BC and PR=AC and also ∠P = ∠A, ∠Q = ∠B and ∠R = ∠C. This is denoted as ∆PQR ≡ ∆ABC .
There are 4 ways by which one can prove that two triangles are congruent.
(i) SSS (Side – Side – Side) Congruence
If the three sides of a triangle are congruent to the three sides of another triangle, then the triangles are congruent.
That is AB = PQ, BC = QR and AC = PR
⇒ ∆ABC ≡ ∆PQR.
(ii) SAS (Side – Angle – Side) Congruence
If two sides and the included angle (the angle between them) of a triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Here, AC = PQ, ∠A = ∠P and AB = PR and hence ∆ACB ≡ ∆PQR.
(iii) ASA (Angle-Side-Angle) Congruence
If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. Here, ∠A = ∠R, CA=PR and ∠C = ∠P and hence ∆ABC ≡ ∆RQP
(iv) RHS (Right Angle – Hypotenuse – Side) Congruence
If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. Here,
∠B = ∠Q = 90°, BC = QR and AC = PR
(right angle) (leg) (hypotenuse)
and hence ∆ABC ≡ ∆PQR.
• Any segment or angle is congruent to itself! This is called Reflexive property
• If two triangles are congruent, then their corresponding parts are congruent. This is called CPCTC (Corresponding parts of Congruent Triangles are Congruent).
• If angles then sides means if two angles are equal in a triangle, then the sides opposite to them are equal.
• If sides then angles means if two sides are equal in a triangle, then the angles opposite to them are equal.
Match the following by their congruence property
Answer: 1. (iv), 2. (iii), 3. (i), 4. (ii)
Find the unknowns in the following figures
(i) Now, from Fig. 5.7 (i), 140º + ∠z = 180º (linear pair)
⇒ ∠z = 180º - 140º = 40º
Also ∠x + ∠z = 70º + ∠z (exterior angle property)
⇒ ∠x = 70º
Also ∠z + ∠y + 70º = 180º (angle sum property in ΔABC)
⇒ 40º + ∠y + 70º = 180º
⇒ ∠y = 180º - 110º = 70º
(ii) Now, from Fig. 5.7(ii), PQ = PR
⇒ ∠Q = ∠R (angles opposite to equal sides are equal)
⇒ ∠x = ∠y
⇒ ∠x+ ∠y + 50º = 180º (angle sum property in ΔPQR)
⇒ ∠2 x = 130º
⇒ ∠x = 65º
⇒ ∠y = 65º
(iii) Now, from Fig. 5.7(iii), in ΔABC
∠A = x (vertically opposite angles)
Similarly ∠B = ∠C = x (Why?)
⇒ ∠A + ∠B + ∠C = 180º (angle sum property in ΔABC)
⇒ 3 x =180º
⇒ x = 60º
⇒ y = 180º − 60º = 120º
Example 5.2 (Illustrating SSS and SAS Congruence)
If ∠E = ∠S and G is the midpoint of ES, prove that ∆GET ≡ ∆GST.
In the figure, DA = DC and BA = BC.
Are the triangles DBA and DBC congruent? Why?
Here AD = CD
AB = CB
DB = DB (common)
ΔDBA = ΔDBC [∵ By SSS congruency]
Also RHS rule also bind here to say their congruency.
Example 5.3 (Illustrating ASA Congruence)
If ∠YTB ≡ ∠YBT and ∠BOY ≡ ∠TRY , prove that ∆BOY ≡ ∆TRY
Example 5.4 (Illustrating RHS Congruence)
If TAP is an isosceles triangle with TA = TP and ∠TSA = 90°.
(i) Is ∆TAS ≡ ∆TPS ? Why?
(ii) Is ∠P = ∠A? Why?
(iii) Is AS = PS? Why?
(i) TA = TP (hypotenuse) and ∠TSA = 90
TS is common (leg)
Hence, by RHS congruence, ∆TAS ≡ ∆TPS
(ii) Given TA = TP
∠P = ∠A (if angles then sides)
(iii) From (i) ∆TAS ≡ ∆TPS ,
SSA and ASS properties are not suﬃcient to prove that two triangles are congruent. This is explained in the given figure. By construction, in triangles ABD and ABC, BC = BD = a. Also, AB and ∠BAZ are common. But AC ≠ AD. So, ∆ABD is not congruent to ∆ABC and so SSA fails.