Tag: 9thmaths 6 th chapter

Example 1: Prove Vertical Opposite Angles are Equal

Question

Given two intersecting lines, prove that vertical opposite angles are equal.

Solution

1. Given: Two lines ABAB and CDCD intersect at OO.
   • Let ∠AOC\angle AOC and ∠BOD\angle BOD be one pair of vertical opposite angles.
   • Let ∠AOD\angle AOD and ∠BOC\angle BOC be the other pair of vertical opposite angles.
2. To Prove:
   ∠AOC=∠BOD\angle AOC = \angle BOD and ∠AOD=∠BOC\angle AOD = \angle BOC.
3. Proof:
   • In △AOC\triangle AOC: ∠AOC+∠BOD=180∘(Linear Pair Axiom).\angle AOC + \angle BOD = 180^\circ \quad \text{(Linear Pair Axiom)}.
   • Similarly, ∠BOD+∠AOD=180∘.\angle BOD + \angle AOD = 180^\circ.
   • From the above equations, ∠AOC=∠BOD.\angle AOC = \angle BOD.
   • Similarly, ∠AOD=∠BOC.\angle AOD = \angle BOC.
4. Conclusion: Vertical opposite angles are equal.

Example 2: Prove Corresponding Angles are Equal

Question

Prove that if a transversal intersects two parallel lines, the corresponding angles are equal.

Solution

1. Given: Two parallel lines l1l_1 and l2l_2, and a transversal tt intersecting them at points PP and QQ.
   • Let ∠1\angle 1 and ∠2\angle 2 be corresponding angles.
2. To Prove:
   ∠1=∠2\angle 1 = \angle 2.
3. Proof:
   • Since l1∥l2l_1 \parallel l_2, and tt is a transversal, alternate interior angles are equal: ∠1=∠3.\angle 1 = \angle 3.
   • But ∠3\angle 3 is vertically opposite to ∠2\angle 2, so: ∠3=∠2.\angle 3 = \angle 2.
   • Therefore: ∠1=∠2.\angle 1 = \angle 2.
4. Conclusion: Corresponding angles are equal when a transversal intersects two parallel lines.

Example 3: Angle Sum Property of a Triangle

Question

Prove that the sum of the angles in a triangle is 180∘180^\circ.

Solution

1. Given: A triangle ABCABC.
   • Extend BCBC to DD.
2. To Prove:
   ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ.
3. Construction:
   • Draw a line DEDE parallel to ABAB through CC.
4. Proof:
   • Since DE∥ABDE \parallel AB, and ACAC is a transversal: ∠A=∠1(Corresponding Angles).\angle A = \angle 1 \quad \text{(Corresponding Angles)}.
   • Similarly, BCBC is a transversal: ∠B=∠2(Corresponding Angles).\angle B = \angle 2 \quad \text{(Corresponding Angles)}.
   • Along the straight line DEDE: ∠1+∠C+∠2=180∘.\angle 1 + \angle C + \angle 2 = 180^\circ.
   • Substituting: ∠A+∠B+∠C=180∘.\angle A + \angle B + \angle C = 180^\circ.
5. Conclusion: The sum of the angles in a triangle is 180∘180^\circ.

Example 4: Exam-Style Problem

Question

In the figure below, AB∥CDAB \parallel CD, and EFEF is a transversal. If ∠AEF=65∘\angle AEF = 65^\circ, find ∠CFE\angle CFE.

Solution

1. Given: AB∥CDAB \parallel CD, EFEF is a transversal, and ∠AEF=65∘\angle AEF = 65^\circ.
2. To Find:
   ∠CFE\angle CFE.
3. Solution:
   • Since AB∥CDAB \parallel CD, and EFEF is a transversal: ∠AEF=∠CFE(Alternate Interior Angles).\angle AEF = \angle CFE \quad \text{(Alternate Interior Angles)}.
   • Substituting: ∠CFE=65∘.\angle CFE = 65^\circ.
4. Conclusion:
   ∠CFE=65∘\angle CFE = 65^\circ.

Practice Problems

1. In a triangle, one angle is 90∘90^\circ, and another is 45∘45^\circ. Find the third angle.
2. Prove that if two lines are parallel, the alternate interior angles are equal.
3. If two angles of a triangle are 50∘50^\circ and 60∘60^\circ, find the third angle.
4. Prove that the exterior angle of a triangle is equal to the sum of its two opposite interior angles.

Would you like diagrams for these problems or additional practice questions?

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