Question

Answer it plz for 10points​

Answer 1

Step-by-step explanation:

Given : Point O is taken inside a rhombus ABCD such that BO = OD.

To prove : AO and OC are in one and the same St. line.

Proof : In ∆s AOD and AOB, we have

AD = AB (given)

AO = AO

OD = OB (given)

∴ ∆ AOD ≌ ∆AOB (SSS axiom of congruence)

∴ ∠1 = ∠2 (cpct)

similarly, ∆DOC ≌ ∆BOC

∴ ∠3 = ∠4

But ∠1 + ∠2 + ∠3 + ∠4 = 360° (∠s round a point)

or 2∠2 + 2∠3 = 360° ( ∵ ∠1 = ∠3 and ∠3 = ∠ 4)

⇛ ∠2 + ∠3 = 180° (proved above)

Therefore, their outer arms AO and BO are in one and the same St. line. (Axiom of linear pair)