Question

Show that the diagonals of a rhombus are perpendicular to each other.

​

Answer 1

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

Consider the rhombus as ABCD,

Let the center point be O

Now in triangle AOD and COD,

OA = OC ( Diagonals of IIgm bisect each other )

OD= OD (common )

AD = CD

Therefore, triangle AOD congruent triangle COD

Thus gives ,

Angle AOD = angle COD (cpct)

= 2 AOD = 180°

= AOD = 90°

So , the diagonals of a rhombus are perpendicular to each other.

_________________________

Hope it helps :)

Answer 2

Answer:

I think it helps you ⬇⬇

Step-by-step explanation:

Given:-ABCD is a rhombus

To prove that:-Diagonals are perpendicular to eachother i.e,AC⊥BD

Solution:-In ΔAOB and ΔAOD

                      AB=AD

                      AO=AO

                      OB=OD

By Side Side Side [S.S.S] congruence rule ΔAOD≅ΔAOD

⇒∠AOB=∠AOD [∵By CPCT] [∵CPCT=Corresponding Parts of Congruent      Triangle]

But ∠AOB+∠AOD=180° [Linear Pair of Angles]

  ⇒∠AOB+∠AOB=180°

                2∠AOB=180°

                  ∠AOB=180°/2

                  ∠AOB=90°

∴Hence AC⊥BD

∴Please mark it as brainlist answer