||Congruence||
Show that the diagonals of a rhombus are perpendicular to each other.
Follow back xD
Answers
Step-by-step explanation:
Proof
(1) ABCD is a rhombus //Given
(2) AB=AD //definition of rhombus
(3) AO=AO //Common side, reflexive property of equality
(4) BO=OD // A rhombus is a parallelogram, a parallelogram’s diagonals bisect each other
(5) △AOD≅△AOB //Side-Side-Side postulate.
(6) ∠AOD ≅ ∠AOB //Corresponding angles in congruent triangles (CPCTC)
(7) AC⊥DB //Linear Pair Perpendicular Theorem
The converse of this is also true: if a parallelogram’s diagonals are perpendicular, it is a rhombus.
Given :- ABCD is a rhombus
AC and BD are diagonals of rhombus intersecting at O.
⇒ To prove :- ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°
⇒ Proof :- All Rhombus are parallelogram, Since all of its sides are equal.
AB = BC = CD = DA ────(1)
The diagonal of a parallelogram bisect each other
Therefore, OB = OD and OA = OC ────(2)
In ∆ BOC and ∆ DOC
BO = OD [ From 2 ]
BC = DC [ From 1 ]
OC = OC [ Common side ]
∆ BOC ≅ ∆ DOC [ By SS congruency criteria ]
∠BOC = ∠DOC [ C.P.C.T ]
∠BOC + ∠DOC = 180° [ Linear pair ]
2∠BOC = 180° [ ∠BOC = ∠DOC ]
∠BOC = 180°/2
∠BOC = 90°
∠BOC = ∠DOC = 90°
Similarly, ∠AOB = ∠AOD = 90°
Hence, ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°
❣
