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

Grade 7ICSE

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

🔑Concepts

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Definition of Congruence: Two triangles are congruent if they are identical in shape and size. When one triangle is placed over the other, they coincide exactly in all respects. The symbol for congruence is ≅\cong. For example, if △ABC≅△PQR\triangle ABC \cong \triangle PQR, it means every side and every angle of △ABC\triangle ABC matches the corresponding side and angle of △PQR\triangle PQR.

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CPCT (Corresponding Parts of Congruent Triangles): This rule states that if two triangles are proved congruent, then their corresponding remaining parts (angles and sides) must also be equal. For instance, if △ABC≅△XYZ\triangle ABC \cong \triangle XYZ is proved by SSS, we can conclude that ∠A=∠X\angle A = \angle X using CPCT.

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SSS Criterion (Side-Side-Side): Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle. Visually, think of two triangles where all three outer boundaries have the same lengths, locking the shape into a fixed, identical configuration.

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SAS Criterion (Side-Angle-Side): 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. The angle must be 'sandwiched' between the two sides. Imagine two sides of fixed length meeting at a specific corner; the distance between the two open ends is then automatically fixed.

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ASA Criterion (Angle-Side-Angle): 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. In a visual sense, if you have a base line (side) of a specific length and you draw two specific angles at its ends, the lines will meet at a unique point to form identical triangles.

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RHS Criterion (Right angle-Hypotenuse-Side): This criterion applies specifically to right-angled triangles. Two right-angled triangles are congruent if the hypotenuse and one side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle. Visually, since the 90∘90^{\circ} angle is already fixed, matching the longest side (hypotenuse) and one leg ensures the third side must be the same by the Pythagorean property.

📐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.