Question

show that diagonals of a rhombus are perpendicular to each other

Answer 1

this is the answer... hope this helps you

Answer 2

Given:-Rhombus ABCD

To prove:-AC bisects BD

Proof:-In ΔAOB and ΔCOB,

AO=OC (Diagonals of a parallelogram bisect each other)

OB=OB (Common)

AB=BC (Sides of a rhombus are equal)

Therefore,

ΔAOB≅ΔCOB (SSS congruency rule)

⇒∠AOB=∠COB (C.P.C.T.)→{1}

Since, AC is a line

∠AOB+∠COB=180°

∠AOB+∠AOB=180°→[From {1}]

2(∠AOB)=180°

∠AOB=180°/2

∠AOB=90°

From{1},

∠COB=∠AOB

∠COB=90°

Also,

∠DOC=∠AOB=90° (Vertically opposite Angles)

∠DOA=∠COB=90°(Vertically opposite Angles)

Since,

∴∠DOC=∠AOB=∠DOA=∠COB=90°

∴AC bisects BD

⇒Hence,diagonals of a rhombus are perpendicular to each other.