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

Grade 9CBSE

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

🔑Concepts

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Congruence of Triangles: Two triangles are congruent if they are copies of each other and when superimposed, they cover each other exactly. This means all corresponding sides and corresponding angles are equal. Visually, if ΔABC≅ΔPQR\Delta ABC \cong \Delta PQR, side ABAB matches PQPQ, BCBC matches QRQR, and ACAC matches PRPR.

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SAS (Side-Angle-Side) Congruence Rule: 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. Visually, look for two sides forming a 'V' shape with the designated angle located at the vertex where the two sides meet.

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ASA (Angle-Side-Angle) and AAS (Angle-Angle-Side) Rules: ASA states that two triangles are congruent if two angles and the included side of one are equal to those of the other. AAS is a variation where congruence is established if any two pairs of angles and one pair of corresponding sides are equal. Visually, in ASA, the equal side is the 'bridge' connecting the two equal angles.

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SSS (Side-Side-Side) Congruence Rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent. Visually, this is represented by matching hash marks on all three pairs of corresponding sides (e.g., one tick mark on ABAB and PQPQ, two on BCBC and QRQR, etc.).

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RHS (Right Angle-Hypotenuse-Side) Congruence Rule: If in two right-angled 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. Visually, look for the 90∘90^\circ square symbol at the vertex, the longest side (hypotenuse) across from it, and one matching leg.

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CPCT (Corresponding Parts of Congruent Triangles): Once two triangles are proved congruent by any of the above criteria (SAS, ASA, SSS, RHS), all their other corresponding parts (sides and angles) are also equal. This is a fundamental logical step used to prove equalities in geometry problems.

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Properties of Isosceles Triangles: Angles opposite to equal sides of an isosceles triangle are equal. Conversely, sides opposite to equal angles of a triangle are equal. Visually, an isosceles triangle appears symmetrical, where the two equal sides form an 'apex' and the two base angles are identical.

📐Formulae

ΔABC≅ΔPQR  ⟹  AB=PQ,BC=QR,AC=PR\Delta ABC \cong \Delta PQR \implies AB=PQ, BC=QR, AC=PR

ΔABC≅ΔPQR  ⟹  ∠A=∠P,∠B=∠Q,∠C=∠R\Delta ABC \cong \Delta PQR \implies \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R

∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ (Angle Sum Property)

If , AB = AC , in , \Delta ABC, , then , \angle C = \angle B$

Area , of , \Delta ABC = Area , of , \Delta PQR , \text{if} , \Delta ABC \cong \Delta PQR$

💡Examples

Problem 1:

In ΔABC\Delta ABC, the bisector ADAD of ∠A\angle A is perpendicular to side BCBC. Show that AB=ACAB = AC and ΔABC\Delta ABC is isosceles.

Solution:

In ΔABD\Delta ABD and ΔACD\Delta ACD:

  1. ∠BAD=∠CAD\angle BAD = \angle CAD (Since ADAD is the bisector of ∠A\angle A)
  2. AD=ADAD = AD (Common side)
  3. ∠ADB=∠ADC=90∘\angle ADB = \angle ADC = 90^\circ (Given AD⊥BCAD \perp BC) Therefore, ΔABD≅ΔACD\Delta ABD \cong \Delta ACD by the ASAASA congruence rule. By CPCT, AB=ACAB = AC. Since two sides of the triangle are equal, ΔABC\Delta ABC is an isosceles triangle.

Explanation:

We use the ASA (Angle-Side-Angle) criterion because we have a common side between two pairs of equal angles. Once congruence is proved, CPCT allows us to conclude the sides are equal.

Problem 2:

PP is a point equidistant from two lines ll and mm intersecting at point AA. Show that the line APAP bisects the angle between them.

Solution:

Let PB⊥lPB \perp l and PC⊥mPC \perp m. Given PB=PCPB = PC (equidistant). In ΔPBA\Delta PBA and ΔPCA\Delta PCA:

  1. ∠PBA=∠PCA=90∘\angle PBA = \angle PCA = 90^\circ (By construction of perpendiculars)
  2. PA=PAPA = PA (Common hypotenuse)
  3. PB=PCPB = PC (Given side) Therefore, ΔPBA≅ΔPCA\Delta PBA \cong \Delta PCA by the RHSRHS congruence rule. Hence, ∠BAP=∠CAP\angle BAP = \angle CAP by CPCT. This proves that APAP bisects the angle between lines ll and mm.

Explanation:

The RHS (Right Angle-Hypotenuse-Side) rule is applied here because we are dealing with distances (perpendiculars) and a shared hypotenuse in right-angled triangles.