Question

7. ABCD is a rhombus. Show that diagonal AC
bisects Z A as well as Z C and diagonal BD
bisects ZB as well as D.​

Answer 1

Step-by-step explanation:

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

⇒AB=BC=CD=DA

⇒AC⊥BD

In Δ ABC

⇒AB=BC

⇒∴∠CAB=∠ACB

⇒AO=OC

∴∠ABO=∠CBO

and similiarly in Δ ADC

⇒AD=DC

⇒AO=OC

∴corresponding angle are equal

⇒∠DAC=∠DCA

and∠CDO=∠ADO

Hence,AC bisect∠Aand∠C and BD bisects∠Band∠D

HENCE, PROVED.......

HOP IT HELPS

Answer 2

Step-by-step explanation:

GIVEN : Rhombus ABCD

TO PROVE : 1 ) AC bisects <A and bisicts <C .

2 ) BD bisects <D and <B

PROOF : In ∆ABC ,

AB=BC ( sides of Rhombus are equal )

so , <4 = <2 ( Angles opposite to equal sides are equal ) ......1

Now , AD||BC ( opposite side of Rhombus are parallel )

and transversal of AC .

<1 = <4 ( Alternate angles ) .......2

From ( 1 ) and ( 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 ) and ( 3 )

<4 =<3

= AC bisects<C

Hence AC bisects <C and <A

Similarly we can prove that BD bisects <B and <D

HENCE PROVED