NCERT Solutions for Class 9 Maths Exercise 8.1 Question 7
Welcome, students! Let’s solve Question 7 from Exercise 8.1. In this problem, we will prove a key property of a rhombus: that its diagonals bisect the angles at the vertices. This proof is a direct application of the SSS congruence rule, based on the definition of a rhombus.
| Given Information | A rhombus ABCD. |
|---|---|
| To Prove | (i) Diagonal AC bisects &&\angle A&& and &&\angle C&&. (ii) Diagonal BD bisects &&\angle B&& and &&\angle D&&. |
| Key Concept Used | Definition of a rhombus (all sides equal) and the SSS congruence rule. |
Question 7: ABCD is a rhombus. Show that diagonal AC bisects &&\angle A&& as well as &&\angle C&& and diagonal BD bisects &&\angle B&& as well as &&\angle D&&.
Detailed Step-by-Step Proof 📝
We will prove the property for each diagonal separately.
Part (i): Show that diagonal AC bisects &&\angle A&& and &&\angle C&&
Let’s consider the two triangles formed by the diagonal AC, which are &&\triangle ABC&& and &&\triangle ADC&&.
In &&\triangle ABC&& and &&\triangle ADC&&:
- &&AB = AD&& (Sides of a rhombus are equal) [S]
- &&BC = DC&& (Sides of a rhombus are equal) [S]
- &&AC = AC&& (Common side) [S]
Therefore, by the SSS (Side-Side-Side) congruence rule, we have:
&&\triangle ABC \cong \triangle ADC&&
By CPCTC (Corresponding Parts of Congruent Triangles are equal), their corresponding angles are equal:
- &&\angle BAC = \angle DAC&&
- &&\angle BCA = \angle DCA&&
This shows that AC bisects both &&\angle A&& and &&\angle C&&.
Part (ii): Show that diagonal BD bisects &&\angle B&& and &&\angle D&&
Similarly, let’s consider the two triangles formed by the diagonal BD, which are &&\triangle ABD&& and &&\triangle CBD&&.
In &&\triangle ABD&& and &&\triangle CBD&&:
- &&AB = CB&& (Sides of a rhombus are equal) [S]
- &&AD = CD&& (Sides of a rhombus are equal) [S]
- &&BD = BD&& (Common side) [S]
Therefore, by the SSS congruence rule, we have:
&&\triangle ABD \cong \triangle CBD&&
By CPCTC, their corresponding angles are equal:
- &&\angle ABD = \angle CBD&&
- &&\angle ADB = \angle CDB&&
This shows that BD bisects both &&\angle B&& and &&\angle D&&.
Converse of Question 6 ✅
This proof is the converse of the property explored in Question 6. Together, they establish a key two-way relationship for parallelograms:
- Question 6: If a diagonal of a parallelogram bisects a vertex angle, then it is a rhombus.
- Question 7: If a parallelogram is a rhombus, then its diagonals bisect the vertex angles.
FAQ (Frequently Asked Questions)
Q: What is the primary property of a rhombus used in this proof?
A: The primary property of a rhombus used here is that all four of its sides are equal in length (&&AB = BC = CD = DA&&). This allows us to use the SSS congruence rule effectively.
Q: What is the SSS congruence rule?
A: The SSS (Side-Side-Side) congruence rule states that if three sides of one triangle are equal to the three corresponding sides of another triangle, then the two triangles are congruent.
Q: How does this question relate to Question 6?
A: This question is the converse of Question 6. In Question 6, we were given a parallelogram where a diagonal bisected an angle and had to prove it was a rhombus. Here, we are given a rhombus and must prove its diagonals bisect the angles. Together, they establish that a parallelogram is a rhombus if and only if its diagonals bisect the vertex angles.
Q: Do the diagonals of a general parallelogram bisect the vertex angles?
A: No. In a general parallelogram (that is not a rhombus or a square), the diagonals do not bisect the vertex angles. This property is specific to rhombuses and squares.
