Question

Show the diagonals of a rhombus are perpendicular to each other​

Answer 1

given: rhombus ABCD

ti prove:Ac perpendicular BD

proof: since ABCDis a rhombus

AB=BC=CD=DA

in triangle AOB and And triangle COB

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

OB=OB(common)

AB=CB(side of rhombus are equal)

triangle AOB=triangle COB(sss congruence rule)

=angle AOB=angle=COB(cpct..(1))

Answer 2

$\large\sf\red{Consider\:the\:rhombus\:ABCD}$

$\huge\sf\green{You\:know\:that,}$

$\large\sf{AB=BC=CD=DA}$

$\large\sf{In\:∆AOD\:and\:∆COD}$

$\large\sf{OA=OC(diagonals\:of\:a\:||gm\:bisect\:each\:other}$

$\large\sf{OD=OD(common)}$

$\large\sf{AD=CD(given)}$

$\large\therefore$$\large\sf{∆AOD\:\cong\:∆COD}$ (By SSS congruence rule)

$\large\sf{\angle{AOD}=\angle{COD}(By\:CPCT)}$

$\large\sf\orange{\angle{AOD}+\angle{COD}=180°(Linear\:pair)}$

$\large\sf\orange{2\angle{AOD}=180°}$

$\large\sf\orange{\angle{AOD}=90°}$

$\small\tt\green{So,\:the\:diagonals\:of\:a\:rhombus\:are\:perpendicular\:to\:each\:other}$