Question

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

PLEASE TRY TO SOLVE THIS QUESTION. OF CLASS 9.

THANK YOU:) ​

Answer 1

Given :-

ABCD is a rhombus.AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D

Solution :-

At first we need to know two things abput a rhombus

(i) Sides of rhombus are equal

(ii) Opposite side of rhombus are equal.

Let,

∠AD = ∠1

∠AB = ∠2

∠BD = ∠3

∠BC = ∠4

Using Property 1

AB = BC

∠2 = ∠4 (i)

AD will be || to BC

∠1 = ∠4

Putting ∠4 = ∠2 from 1

∠1 = ∠2

And

AB will be || to CD

∠2 = ∠3

Putting ∠2 = ∠4 from 1

∠4 = ∠3

Therefore,

AC bisects ∠A and ∠C

Similarly

BD will bisect ∠B and ∠D

Answer 2

Given :-

• ABCD is a rhombus.

To Show :-

• AC bisect ∠A and ∠C
• BD bisect ∠B and ∠D

Solution and concept :-

To show that Diagonal bisect the angles of rhombus , at first we have to draw a rhombus then add Vertex A to C and Vertex B to D such that AC and BD become the bisector of angles of rhombus. As we know that in rhombus all sides are equal and it's opposite sides are also equal. To show diagonal bisect its angles , at first we have to prove congruency as per the figure. Concept, remember bisector always the whole angles bisect into two equal parts:

As per the construction figures :-

∆ADC and ∆BAC

⟼ AD = BC ( AD||BC)

⟼ DC = AB ( DC||AB)

⟼ ∠D = ∠B ( opposite angles are equal)

∆ADC ≅ ∆BAC (by S.A.S criterion of congruency)

⟼ ∠ABC = ∠ABD + ∠CBD ( by CPCTC)

⟼ ∠ABD = ∠CBD ( by CPCTC)

∆DCB and ∆BAD

⟼ AD = BC ( AD||BC)

⟼ DC = AB ( DC||AB)

⟼ ∠A = ∠C ( opposite angles are equal)

∆DCB ≅ ∆BAD (by S.A.S criterion of congruency)

⟼ ∠DAB = ∠DAC + ∠BAC (by CPCTC)

⟼ ∠DAC = ∠BAC ( by CPCTC)

From above statement :-

• AD and BD Bisect angles
• Hence, proved