Question

Proof that the diagonals of a rhombus bisect each other at their points of intersection

Answer 1

Step-by-step explanation:

Given:  ABCD is a rhombus in which AB = BC = CD = AD.

RTP: OD = OB and OA = OC

Proof :  

From ΔAOB and ΔAOD

AB = AD     (From given)

AO = AO     (Common side)

Since diagonals in rhombus perpendicular to each other

∠AOB =∠AOD =  90°  

∴ ΔAOB ≅ ΔAOD         (By RHS congruence)

So. OD = OB ---- (1)    (By CPCT)

Similarly  

From ΔAOB and ΔBOC

We can prove OA = OC  ----- (2)

Therefore, ,from (1) and (2) we can say that diagonals of a rhombus bisect each other.

Answer 2

Answer:

..

Step-by-step explanation:

Answer;-

Given= ABCD is a rhombus in which AB=BC=CD=AD.

RTP: OD=OB and OA=OC

Proof:

From ∆AOB and ∆AOD

AB=AD (Given)

AO=AO (Common)

Since, diagonal in rhimbus is perpendicular to each other

ᒪAOB=ᒪBOD =90°

Therefore, ∆AOB = ∆BOC