Question

Prove that in a rhombus, the diagonals are perpendicular to each other.

Answer 1

Given: $ABCD$ is a rhombus; $AC$ and $BD$ intersect at $E$.

To prove: $AC\perp BD$

In rhombus, $ABCD$

$AC$ and $BD$ intersect each Other at $E$ (Given)

In $\Delta ABE$ and $\Delta ADE$

$AB=AD$ (sides of a rhombus)

$BE=DE$ (Diagonals bisect each other)

$AE$ common

$\therefore △ABE=△ADE$

$⸫AEB+AED$ (CPCT)

$\angle AEB+\angle AED={180}^0$ (Linear pair)

$\Rightarrow AEB=AED=\frac{180}{2}={90}^0$

Hence, $AC$ and $BD$ are perpendicular to each other.