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Triangles - Congruence Theorems - Solve multi-step geometric problems based on triangle theorems

Grade 9CBSE

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

🔑Concepts

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The Side-Angle-Side (SAS) Congruence Rule states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent. This is a primary tool for proving equality of parts in geometric proofs using CPCT (Corresponding Parts of Congruent Triangles).

Two triangles ABC and PQR illustrating SAS congruence with marked equal sides AB=PQ and BC=QR.
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The Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) rules are used when two angles and one side are given. In ASA, the side must be included between the angles, whereas in AAS, the side is opposite one of the angles. Both criteria imply triangle congruence.

A triangle with two base angles and the included base side marked to represent the ASA congruence criterion.
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The Right Angle-Hypotenuse-Side (RHS) Congruence Rule is specific to right-angled triangles. It states that if the hypotenuse and one side of one right triangle are equal to the corresponding hypotenuse and side of another right triangle, the triangles are congruent.

A right-angled triangle with the right angle, hypotenuse, and one leg identified for the RHS rule.
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In an Isosceles Triangle, angles opposite to equal sides are equal. Conversely, sides opposite to equal angles of a triangle are equal. This property is frequently used in multi-step problems to transition from side equalities to angle equalities and vice versa.

📐Formulae

In ΔABC,AB+BC>AC\text{In } \Delta ABC, AB + BC > AC

In ΔABC,BC+AC>AB\text{In } \Delta ABC, BC + AC > AB

In ΔABC,AB+AC>BC\text{In } \Delta ABC, AB + AC > BC

|AB - BC| < AC

If ∠A>∠B>∠C, then BC>AC>AB\text{If } \angle A > \angle B > \angle C, \text{ then } BC > AC > AB

💡Examples

Problem 1:

Is it possible to construct a triangle with sides of lengths 5 cm5\text{ cm}, 8 cm8\text{ cm}, and 15 cm15\text{ cm}?

Solution:

  1. According to the Triangle Inequality Theorem, the sum of any two sides must be greater than the third side.
  2. Let a=5a = 5, b=8b = 8, and c=15c = 15.
  3. Check the sum of the two smaller sides: a+b=5+8=13a + b = 5 + 8 = 13.
  4. Compare this sum to the third side: 13<1513 < 15.
  5. Since the sum of two sides is not greater than the third side (13≯1513 \ngtr 15), a triangle cannot be formed.

Explanation:

To check if a triangle exists, you only need to verify if the sum of the two shortest sides is strictly greater than the longest side.

Problem 2:

In ΔPQR\Delta PQR, if ∠P=45∘\angle P = 45^\circ and ∠Q=65∘\angle Q = 65^\circ, determine which side of the triangle is the longest and which is the shortest.

Solution:

  1. First, find the third angle using the Angle Sum Property: ∠R=180∘−(∠P+∠Q)=180∘−(45∘+65∘)=180∘−110∘=70∘\angle R = 180^\circ - (\angle P + \angle Q) = 180^\circ - (45^\circ + 65^\circ) = 180^\circ - 110^\circ = 70^\circ.
  2. Compare the angle measures: 45∘<65∘<70∘45^\circ < 65^\circ < 70^\circ, so ∠P<∠Q<∠R\angle P < \angle Q < \angle R.
  3. Use the property that the side opposite the larger angle is longer:
    • Side opposite to ∠R\angle R (70∘70^\circ) is PQPQ.
    • Side opposite to ∠Q\angle Q (65∘65^\circ) is PRPR.
    • Side opposite to ∠P\angle P (45∘45^\circ) is QRQR.
  4. Therefore, QR<PR<PQQR < PR < PQ. The longest side is PQPQ and the shortest side is QRQR.

Explanation:

The relative lengths of the sides of a triangle are determined by the measures of the angles opposite to them. Larger angles face longer sides.

Problem 3:

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

Geometry diagram showing overlapping triangles ABC and ADE sharing vertex A.

Solution:

  1. Given: AB=ADAB = AD, AC=AEAC = AE and ∠BAD=∠EAC\angle BAD = \angle EAC.
  2. Add ∠DAC\angle DAC to both sides of the angle equation: ∠BAD+∠DAC=∠EAC+∠DAC\angle BAD + \angle DAC = \angle EAC + \angle DAC ∠BAC=∠DAE\angle BAC = \angle DAE
  3. Consider ΔABC\Delta ABC and ΔADE\Delta ADE:
  • AB=ADAB = AD (Given)
  • ∠BAC=∠DAE\angle BAC = \angle DAE (Proved above)
  • AC=AEAC = AE (Given)
  1. Therefore, ΔABC≅ΔADE\Delta ABC \cong \Delta ADE by SASSAS congruence rule.
  2. Hence, BC=DEBC = DE by CPCTCPCT.

Explanation:

To prove BC=DEBC = DE, we identify triangles ΔABC\Delta ABC and ΔADE\Delta ADE that contain these segments. We use the given angle equality and add the common angle ∠DAC\angle DAC to establish the equality of the included angles, allowing the use of the SAS rule.

Problem 4:

In ΔABC\Delta ABC, the perpendicular bisector ADAD of side BCBC is drawn. Show that ΔABC\Delta ABC is an isosceles triangle in which AB=ACAB = AC.

Triangle ABC with altitude AD drawn from A to BC, marking BD=DC and angle ADC as 90 degrees.

Solution:

  1. In ΔABD\Delta ABD and ΔACD\Delta ACD:
  • BD=CDBD = CD (Since ADAD is the bisector of BCBC)
  • ∠ADB=∠ADC=90∘\angle ADB = \angle ADC = 90^\circ (Since ADAD is perpendicular to BCBC)
  • AD=ADAD = AD (Common side)
  1. By SAS congruence criterion, ΔABD≅ΔACD\Delta ABD \cong \Delta ACD.
  2. Therefore, AB=ACAB = AC by CPCTCPCT.
  3. Since two sides are equal, ΔABC\Delta ABC is an isosceles triangle.

Explanation:

By splitting the triangle into two right-angled triangles using the perpendicular bisector, we can prove those two triangles are congruent using SAS. Once proven, the outer sides AB and AC must be equal by CPCT.