Question

$\bf \huge {\underline {\underline \red{QuEsTiOn}}}$

Prove that the diagonals of a rhombus bisect each other at right angles.​

Answer 1

Answer:

Let ABCD is a rhombus

=> AB=BC=CD=DA

Adjacent sides are equal in rhombus

In ∆ AOD and ∆COD

=>OA=OC [ Diagonals of a rhombus bisect each other]

=>OD=OD [ Common side]

=>AD=AD

=>∆AOD ~= ∆COD [By SSS congruence rule]

=> Angle AOD = Angle COD ( CPCT)

=>Angle AOD + Angle COD= 180° ( linear pair)

=> 2 angle AOD= 180°

=> Angle AOD= 90°

Hope it's helpful to you

Answer 2

$\huge\underline\mathfrak\red{Answer⤵}$

To prove,

Diagonals of rhombus bisect each other at 90°

Proof,

Let the rhombus be named as ABCD

It says that,

AB=BC=CD=DA

As all sides of a rhombus are equal.

In ∆AOD and ∆COD,

OA=OC (diagonals of a rhombus bisect each other)

OD=OD(common side)

AD=AD

Therefore,

∆AOD=∆COD by SSS(side side side)congruency.

Angle AOD=AOC(common parts of congruent traingle)(CPCT)

Therefore,

Angle AOD+Angle COD= Linear pair(180°)

So,

2AOD=180°

AOD=90°

COD=90°

Hence,proved

Hope it helps..

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