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

${\boxed{\bold\red{ANSWER}}}$

[In rhombus all side are equal and diagonols are different but intersect at 90°]

⇒AB=BC=CD=DA

⇒AC⊥BD

In ΔABC

⇒AB=BC

⇒∴∠CAB=∠ACB

⇒AO=OC

∴∠ABO=∠CBO

and simillarly in ΔADC

⇒AD=DC

⇒AO=OC

∴corresponding angle are equal

⇒∠DAC=∠DCA

and ∠CDO=∠ADO

Hence, AC bisect ∠A and ∠C and BD bisects ∠B and ∠D

Answer 2

Step-by-step explanation:

Given,

Rhombus ABCD

To Prove,

(i)AC bisects ∠A ,i.e. ∠ 1 = ∠2

BD bisects ∠B,i.e. ∠ 3 = ∠ 4

(ii) BD bisects ∠ B & ∠ D

Proof,

In ∆ABC ,

AB = BC [ sides of the rhombus]

so, ∠ 4 = ∠ 2 [ angle opposite to equal sides are equal] —(1)

Now, AD || BC [ opposite sides of rhombus are parallel]

and Transversal AC

∠ 1 = ∠ 4 [ alternate angles]—(2)

from (1) &(2)

∠ 1 = ∠ 2

=> AC bisects ∠ A

Now, AB || DC [opposite sides of rhombus are parallel]

and transversal AC

∠ 2 = ∠ 3 [ alternate angles]—(3)

from (1) &(3)

∠ 4 = ∠ 3

=> AC bisects ∠ C

Hence,

AC bisects ∠ A and ∠ C

similarly we can prove that BD bisects ∠ B & ∠ D

Hence proved