krit.club logo

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

Grade 9ICSE

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

🔑Concepts

•

The Side-Angle-Side (SAS) Congruency Rule: If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent. This ensures that the triangles are identical in shape and size.

Two triangles ABC and DEF showing equal corresponding sides AB=DE, BC=EF and equal included angles B and E.
•

Properties of an Isosceles Triangle: In a triangle where two sides are equal, the angles opposite to those sides are also equal. Conversely, if two angles are equal, the sides opposite them are equal. The altitude from the vertex angle bisects the base.

Isosceles triangle ABC with AB=AC and altitude AD from A to BC.
•

Triangle Inequalities: In any triangle, the side opposite to the greater angle is longer, and the angle opposite to the longer side is greater. Furthermore, the sum of any two sides must be strictly greater than the third side.

•

RHS (Right-Angle Hypotenuse Side) Rule: Two right-angled triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.

📐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.

Problem 3:

In the given figure, AB=ACAB = AC and DD is a point in the interior of ΔABC\Delta ABC such that DB=DCDB = DC. Prove that ΔABD≅ΔACD\Delta ABD \cong \Delta ACD and ∠ABD=∠ACD\angle ABD = \angle ACD.

Triangle ABC with an interior point D. Lines AD, BD, and CD are drawn.

Solution:

In ΔABD\Delta ABD and ΔACD\Delta ACD:

  1. AB=ACAB = AC (Given)
  2. DB=DCDB = DC (Given)
  3. AD=ADAD = AD (Common side) By SSS Congruency Rule, ΔABD≅ΔACD\Delta ABD \cong \Delta ACD. Since the triangles are congruent, their corresponding parts are equal. Therefore, ∠ABD=∠ACD\angle ABD = \angle ACD (CPCT).

Explanation:

We use the SSS (Side-Side-Side) rule because all three pairs of corresponding sides are given or shared. CPCT stands for 'Corresponding Parts of Congruent Triangles'.

Problem 4:

In ΔXYZ\Delta XYZ, side XYXY is produced to WW. If XZ=YZXZ = YZ and ∠XZY=40∘\angle XZY = 40^{\circ}, find the measure of exterior angle ∠ZYW\angle ZYW.

Triangle XYZ with side XY extended to W. Angle Z is 40 degrees.

Solution:

In ΔXYZ\Delta XYZ, given XZ=YZXZ = YZ. Therefore, ∠YXZ=∠XYZ\angle YXZ = \angle XYZ (Angles opposite to equal sides). Let ∠YXZ=∠XYZ=x\angle YXZ = \angle XYZ = x. By Angle Sum Property: x+x+40∘=180∘x + x + 40^{\circ} = 180^{\circ} 2x=140∘  ⟹  x=70∘2x = 140^{\circ} \implies x = 70^{\circ}. Since XYWXYW is a straight line, ∠XYZ+∠ZYW=180∘\angle XYZ + \angle ZYW = 180^{\circ} (Linear pair). 70∘+∠ZYW=180∘  ⟹  ∠ZYW=110∘70^{\circ} + \angle ZYW = 180^{\circ} \implies \angle ZYW = 110^{\circ}.

Explanation:

First, use isosceles triangle properties to find the base angles, then use the property of linear pairs to find the exterior angle.