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Triangles - Congruence Theorems - Use triangle congruence criteria (SSS, SAS, ASA, RHS, AAS) in proofs

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. In △ABC≅△PQR\triangle ABC \cong \triangle PQR, the corresponding parts (sides and angles) are equal. This is known as CPCT (Corresponding Parts of Congruent Triangles).

Two identical triangles ABC and PQR illustrating congruence.
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SAS (Side-Angle-Side): 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.

Diagram showing the included angle between two sides for SAS criteria.
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ASA (Angle-Side-Angle) and AAS (Angle-Angle-Side): Two triangles are congruent if two angles and the included side of one are equal to those of the other (ASA). If any two pairs of angles and one pair of corresponding sides are equal, the triangles are congruent (AAS).

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SSS (Side-Side-Side): If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

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RHS (Right Angle-Hypotenuse-Side): 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.

Right-angled triangle showing the hypotenuse and one side for RHS congruence.

📐Formulae

△ABC≅△PQR  ⟹  AB=PQ,BC=QR,AC=PR\triangle ABC \cong \triangle PQR \implies AB=PQ, BC=QR, AC=PR

△ABC≅△PQR  ⟹  ∠A=∠P,∠B=∠Q,∠C=∠R\triangle ABC \cong \triangle PQR \implies \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R

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

Triangle Inequality: AB+BC>AC\text{Triangle Inequality: } AB + BC > AC

In △ABC, if AB=AC  ⟺  ∠C=∠B\text{In } \triangle ABC, \text{ if } AB = AC \iff \angle C = \angle B

💡Examples

Problem 1:

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

Solution:

  1. In △ABD\triangle ABD and △ACD\triangle ACD:
  2. ∠BAD=∠CAD\angle BAD = \angle CAD (Given that ADAD bisects ∠A\angle A)
  3. AD=ADAD = AD (Common side to both triangles)
  4. ∠ADB=∠ADC=90∘\angle ADB = \angle ADC = 90^\circ (Given AD⊥BCAD \perp BC)
  5. Therefore, △ABD≅△ACD\triangle ABD \cong \triangle ACD by the ASA congruence rule.
  6. So, AB=ACAB = AC by CPCT.
  7. Since two sides of △ABC\triangle ABC are equal, it is an isosceles triangle.

Explanation:

We use the properties of the angle bisector and the perpendicularity to establish two angles and a shared side, satisfying the ASA criteria. Once congruence is proved, CPCT allows us to equate the main sides of the triangle.

Problem 2:

Line segment ABAB is parallel to another line segment CDCD. OO is the mid-point of ADAD. Show that △AOB≅△DOC\triangle AOB \cong \triangle DOC and OO is also the mid-point of BCBC.

Solution:

  1. Consider △AOB\triangle AOB and △DOC\triangle DOC:
  2. ∠OAB=∠ODC\angle OAB = \angle ODC (Alternate interior angles as AB∥CDAB \parallel CD and ADAD is the transversal)
  3. OA=ODOA = OD (Given OO is the mid-point of ADAD)
  4. ∠AOB=∠DOC\angle AOB = \angle DOC (Vertically opposite angles)
  5. Thus, △AOB≅△DOC\triangle AOB \cong \triangle DOC by the ASA congruence rule.
  6. Consequently, OB=OCOB = OC by CPCT.
  7. Since OB=OCOB = OC, OO is the mid-point of BCBC.

Explanation:

The parallel lines provide equal alternate interior angles. Combined with the midpoint definition and vertically opposite angles, we satisfy the ASA rule. CPCT is then used to prove the second part of the problem regarding the other midpoint.

Problem 3:

In the given figure, AC=AEAC = AE, AB=ADAB = AD and ∠BAD=∠EAC\angle BAD = \angle EAC. Show that BC=DEBC = DE.

Geometric figure with overlapping triangles sharing vertex A.

Solution:

  1. Given: ∠BAD=∠EAC\angle BAD = \angle EAC.
  2. Add ∠DAC\angle DAC to both sides: ∠BAD+∠DAC=∠EAC+∠DAC\angle BAD + \angle DAC = \angle EAC + \angle DAC. Therefore, ∠BAC=∠DAE\angle BAC = \angle DAE.
  3. In △ABC\triangle ABC and △ADE\triangle ADE:
    • AB=ADAB = AD (Given)
    • ∠BAC=∠DAE\angle BAC = \angle DAE (Proved above)
    • AC=AEAC = AE (Given)
  4. By SAS congruence criterion, △ABC≅△ADE\triangle ABC \cong \triangle ADE.
  5. Hence, BC=DEBC = DE by CPCT.

Explanation:

The key is to identify the correct pair of triangles (△ABC\triangle ABC and △ADE\triangle ADE) and prove the equality of the included angle ∠BAC\angle BAC and ∠DAE\angle DAE by adding a common angle to the given equal angles.

Problem 4:

ABCDABCD is a quadrilateral in which AD=BCAD = BC and ∠DAB=∠CBA\angle DAB = \angle CBA. Prove that △ABD≅△BAC\triangle ABD \cong \triangle BAC.

Quadrilateral ABCD with diagonals AC and BD.

Solution:

  1. Consider △ABD\triangle ABD and △BAC\triangle BAC.
  2. AD=BCAD = BC (Given).
  3. ∠DAB=∠CBA\angle DAB = \angle CBA (Given).
  4. AB=BAAB = BA (Common side).
  5. Therefore, △ABD≅△BAC\triangle ABD \cong \triangle BAC by SAS congruence criterion.

Explanation:

We use the SAS criterion by identifying two sides and the included angle that are common or given as equal in both triangles sharing the base ABAB.