Question

ABCD is a rhombus whose diagonals intersect at O. Show that AAOB = ACOD.​

Answer 1

Answer:

Diagram:-

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Given:-

ABCD is a rhombus whose diagonals intersect at O.

To prove:-

$\sf \triangle AOB=\triangle COD$

Proof:-

• As it is a Rhombus ,

$\sf AB=BC=AD=BD\dots(1)$

• And,As we know that in a Rhombus diagonals bisect each other and are equal in length

$\sf AO=CO,BO=DO\dots (2)$

• And also the Diagonals form right angle while bisecting

Now in $\sf \triangle AOB\:and\:\triangle COD ,$

$\because \begin{cases}\sf AB=CD \\ \sf AO=CO \\ \sf BO=DO\end {cases}$

$\therefore\sf \triangle AOB\cong \triangle COD\left [Side-Side-Side (SSS)\right]\qquad {}_{(proved)}$

We can also prove it using another way

$\sf in \triangle AOB\:and \:\triangle COD,$

$\because \begin{cases}\sf AO=CO \\ \sf BO=DO \\ \sf m\angle AOB=m\angle COD (90°)\end {cases}$

$\therefore\sf \triangle AOB\cong \triangle COD\left[Side-Angle-Side (SAS)\right]\qquad {}_{(proved)}$