ABCD is a rhombus. Show that diagonal AC bisects `/_A`as well as `/_C`and diagonal BD bisects `/_B`as well as `/_Ddot`

Explanation

In`/_\ABD \and\ /_\BCD`
`AB=BC` (Sides of a rhombus)
`AD=CD`(Sides of a rhombus)
`AC=AC` (Common side)
`/_\ABD~=/_\BCD` (SSS test of congruence)
`/_ABD=/_CBD` (C.P.C.T.)
`/_ADB=/_CDB` (C.P.C.T.)
So, Diagonal BD bisects `/_B \and\/_D` of the rhombus ABCD
Similarly we can prove,`/_\ABC~=/_\ADC`
`
/_BAC=/_DAC`(C.P.C.T.)
`/_BCA=/_DCA` (C.P.C.T)
So, Diagonal AC bisects `/_A \and\ /_C` of the rhombus ABCD.