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Geometry - Triangles (Congruency, Isosceles triangle properties, Inequalities)

Grade 9ICSE

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 identical in shape and size. When one triangle is placed over the other, they cover each other exactly. This is represented by the symbol ≅\cong. If ΔABC≅ΔPQR\Delta ABC \cong \Delta PQR, then all corresponding sides (AB=PQ,BC=QR,AC=PRAB=PQ, BC=QR, AC=PR) and corresponding angles (∠A=∠P,∠B=∠Q,∠C=∠R\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R) are equal.

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SAS and ASA Criteria: The Side-Angle-Side (SAS) rule states that two triangles are congruent if two sides and the included angle of one are equal to the corresponding parts of the other. The Angle-Side-Angle (ASA) rule states that if two angles and the included side of one triangle are equal to the corresponding parts of another, they are congruent. Visually, the 'included' part must be physically located between the other two matching components.

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SSS and RHS Criteria: Side-Side-Side (SSS) congruence occurs when all three sides of one triangle match the three sides of another. The Right-angle Hypotenuse Side (RHS) rule is specific to right-angled triangles; they are congruent if the hypotenuse and one side of one triangle equal the hypotenuse and one side of the other. Imagine two right-angled triangles where the long diagonal slope and one leg are identical.

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Isosceles Triangle Properties: An isosceles triangle has at least two equal sides. A key theorem states that angles opposite to equal sides of an isosceles triangle are equal. Conversely, if two angles of a triangle are equal, the sides opposite them are also equal. Visually, an isosceles triangle appears symmetrical, and an altitude drawn from the vertex angle to the base bisects the base at a 90∘90^{\circ} angle.

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Triangle Inequality Theorem: In any triangle, the sum of the lengths of any two sides is always greater than the length of the third side (AB+BC>ACAB + BC > AC). If this condition isn't met, the two shorter segments cannot reach each other to form a vertex, and the triangle cannot be closed.

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Side-Angle Relationship: In a triangle, if two sides are unequal, the angle opposite the longer side is greater than the angle opposite the shorter side. Similarly, the side opposite the largest angle is always the longest side. Imagine a hinge: as you open an angle wider, the side connecting the two ends must stretch further.

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Exterior Angle Property: An exterior angle of a triangle is formed by extending one of its sides. This exterior angle is equal to the sum of the two interior opposite angles. Visually, if you extend side BCBC to point DD, the angle ∠ACD\angle ACD is the sum of ∠A\angle A and ∠B\angle B.

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CPCTC (Corresponding Parts of Congruent Triangles are Congruent): Once two triangles are proven congruent using criteria like SAS or SSS, we can conclude that all their other corresponding parts (sides and angles) are equal. This is the most common method used in geometric proofs to find unknown lengths or angles.

📐Formulae

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

Triangle Inequality: a+b>ca + b > c, b+c>ab + c > a, and c+a>bc + a > b

Difference Inequality: ∣a−b∣<c|a - b| < c

Exterior Angle Theorem: Ext. ∠ACD=∠CAB+∠ABC\text{Ext. } \angle ACD = \angle CAB + \angle ABC

Isosceles Property: If AB=AC  ⟺  ∠B=∠CAB = AC \iff \angle B = \angle C

💡Examples

Problem 1:

In ΔABC\Delta ABC, AB=ACAB = AC and ADAD is the bisector of ∠A\angle A meeting BCBC at DD. Prove that ΔABD≅ΔACD\Delta ABD \cong \Delta ACD and hence show BD=CDBD = CD.

Solution:

  1. In ΔABD\Delta ABD and ΔACD\Delta ACD:
    • AB=ACAB = AC (Given)
    • ∠BAD=∠CAD\angle BAD = \angle CAD (Since ADAD bisects ∠A\angle A)
    • AD=ADAD = AD (Common side)
  2. Therefore, ΔABD≅ΔACD\Delta ABD \cong \Delta ACD by the SASSAS (Side-Angle-Side) criterion.
  3. Since the triangles are congruent, BD=CDBD = CD by CPCTCCPCTC (Corresponding Parts of Congruent Triangles are Congruent).

Explanation:

We use the given side equality and the angle bisector property to identify two sides and an included angle that match, satisfying the SAS rule. CPCTC then allows us to conclude the remaining sides are equal.

Problem 2:

In ΔPQR\Delta PQR, if ∠P=80∘\angle P = 80^{\circ} and ∠Q=60∘\angle Q = 60^{\circ}, identify the longest and shortest sides of the triangle.

Solution:

  1. First, find the third angle ∠R\angle R using the Angle Sum Property: ∠R=180∘−(∠P+∠Q)\angle R = 180^{\circ} - (\angle P + \angle Q) ∠R=180∘−(80∘+60∘)=180∘−140∘=40∘\angle R = 180^{\circ} - (80^{\circ} + 60^{\circ}) = 180^{\circ} - 140^{\circ} = 40^{\circ}
  2. Compare the angles: ∠P(80∘)>∠Q(60∘)>∠R(40∘)\angle P (80^{\circ}) > \angle Q (60^{\circ}) > \angle R (40^{\circ}).
  3. According to the Side-Angle relationship:
    • The side opposite the largest angle (∠P\angle P) is QRQR. So, QRQR is the longest side.
    • The side opposite the smallest angle (∠R\angle R) is PQPQ. So, PQPQ is the shortest side.

Explanation:

The problem applies the Triangle Inequality/Relationship concept where side length is directly proportional to the size of the opposite angle. We must calculate all interior angles before comparing.