krit.club logo

Geometry - Congruence of Triangles: Criteria (SSS, SAS, ASA, RHS)

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Congruence of Triangles: Two triangles are congruent if they are identical in shape and size. This means their corresponding sides and corresponding angles are equal. We use the symbol ≅\cong to denote congruence. For example, △ABC≅△PQR\triangle ABC \cong \triangle PQR means AB=PQAB = PQ, BC=QRBC = QR, AC=PRAC = PR, ∠A=∠P\angle A = \angle P, ∠B=∠Q\angle B = \angle Q, and ∠C=∠R\angle C = \angle R.

Two identical triangles ABC and PQR illustrating congruence.
•

SSS (Side-Side-Side) Criterion: Two triangles are congruent if the three sides of one triangle are equal to the three corresponding sides of the other triangle.

Diagram showing SSS congruence with tick marks on equal sides.
•

SAS (Side-Angle-Side) Criterion: Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.

Diagram showing SAS congruence with an included angle.
•

ASA (Angle-Side-Angle) Criterion: Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

Diagram showing ASA congruence with two angles and the side between them.
•

RHS (Right Angle-Hypotenuse-Side) Criterion: Two right-angled triangles are congruent if the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle.

Diagram showing RHS congruence in right-angled triangles.

📐Formulae

Congruence Symbol: △ABC≅△DEF\triangle ABC \cong \triangle DEF

SSS Condition: AB=DE,BC=EF,AC=DFAB = DE, BC = EF, AC = DF

SAS Condition: AB=DE,∠B=∠E,BC=EFAB = DE, \angle B = \angle E, BC = EF

ASA Condition: ∠A=∠D,AB=DE,∠B=∠E\angle A = \angle D, AB = DE, \angle B = \angle E

RHS Condition: ∠B=∠E=90∘,AC=DF(hypotenuse),AB=DE(side)\angle B = \angle E = 90^{\circ}, AC = DF (hypotenuse), AB = DE (side)

Angle Sum Property: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^{\circ}

💡Examples

Problem 1:

In △ABC\triangle ABC and △PQR\triangle PQR, it is given that AB=5 cmAB = 5\text{ cm}, BC=6 cmBC = 6\text{ cm}, and ∠B=40∘\angle B = 40^{\circ}. In △PQR\triangle PQR, PQ=5 cmPQ = 5\text{ cm}, QR=6 cmQR = 6\text{ cm}, and ∠Q=40∘\angle Q = 40^{\circ}. Are the triangles congruent? If yes, state the criterion.

Solution:

  1. Compare the given components:
    • Side ABAB of △ABC\triangle ABC = Side PQPQ of △PQR=5 cm\triangle PQR = 5\text{ cm}
    • Side BCBC of △ABC\triangle ABC = Side QRQR of △PQR=6 cm\triangle PQR = 6\text{ cm}
    • Included ∠B\angle B of △ABC\triangle ABC = Included ∠Q\angle Q of △PQR=40∘\triangle PQR = 40^{\circ}
  2. Since two sides and the included angle of △ABC\triangle ABC are equal to the corresponding parts of △PQR\triangle PQR, the triangles satisfy the SAS criterion.
  3. Therefore, △ABC≅△PQR\triangle ABC \cong \triangle PQR by SAS congruence criterion.

Explanation:

We identify that the given angle is the 'included angle' (the angle formed between the two known sides). Since the side-angle-side sequence matches in both triangles, they are congruent by SAS.

Problem 2:

In an isosceles △XYZ\triangle XYZ, XY=XZXY = XZ. If XMXM is the perpendicular dropped from XX to the base YZYZ, prove that △XMY≅△XMZ\triangle XMY \cong \triangle XMZ.

Solution:

In △XMY\triangle XMY and △XMZ\triangle XMZ:

  1. ∠XMY=∠XMZ=90∘\angle XMY = \angle XMZ = 90^{\circ} (Given that XM⊥YZXM \perp YZ)
  2. Hypotenuse XY=XY = Hypotenuse XZXZ (Given as an isosceles triangle)
  3. Side XM=XM = Side XMXM (Common side to both triangles)
  4. By the RHS criterion, △XMY≅△XMZ\triangle XMY \cong \triangle XMZ.

Explanation:

To prove congruence in right-angled triangles formed by an altitude, we look for the Right angle, the Hypotenuse, and one common or given Side. Here, XMXM is common, and the slanted sides of the isosceles triangle act as equal hypotenuses.

Problem 3:

In the given figure, ADAD bisects ∠BAC\angle BAC and AD⊥BCAD \perp BC. Prove that △ABD≅△ACD\triangle ABD \cong \triangle ACD using the ASA criterion.

Triangle ABC with altitude AD from A to BC.

Solution:

In △ABD\triangle ABD and △ACD\triangle ACD:

  1. ∠BAD=∠CAD\angle BAD = \angle CAD (Given, ADAD bisects ∠BAC\angle BAC)
  2. AD=ADAD = AD (Common side)
  3. ∠ADB=∠ADC=90∘\angle ADB = \angle ADC = 90^{\circ} (Given, AD⊥BCAD \perp BC) Therefore, △ABD≅△ACD\triangle ABD \cong \triangle ACD by ASA Congruence Criterion.

Explanation:

To use ASA, we identify two angles and the side between them. Here, the side ADAD is shared, and the angles on either side of this segment are equal due to the bisector and perpendicularity conditions.

Problem 4:

In the figure, OO is the mid-point of ABAB and CDCD. Prove that △AOC≅△BOD\triangle AOC \cong \triangle BOD and hence AC=BDAC = BD.

Two intersecting lines AB and CD forming triangles AOC and BOD.

Solution:

In △AOC\triangle AOC and △BOD\triangle BOD:

  1. OA=OBOA = OB (OO is the mid-point of ABAB)
  2. ∠AOC=∠BOD\angle AOC = \angle BOD (Vertically opposite angles)
  3. OC=ODOC = OD (OO is the mid-point of CDCD) Therefore, △AOC≅△BOD\triangle AOC \cong \triangle BOD by SAS Congruence Criterion. By CPCT (Corresponding Parts of Congruent Triangles), AC=BDAC = BD.

Explanation:

Since OO is the midpoint of both segments, the segments are divided into equal halves. The angles formed at the intersection OO are vertically opposite and thus equal, satisfying the Side-Angle-Side condition.

Congruence of Triangles: Criteria (SSS, SAS, ASA, RHS) Class 7 Notes & Examples