Show that bisectors of angles of the parallelogram forms a rectanglepl

1.	Show that bisectors of angles of the parallelogram forms a rectangleplease answer
Answer» ANSWER: Solution: $\sf{Let \ \angle DAB \ be \ 2x \ and \ \angle CBA}$ $\sf{be \ 2y}$ $\sf{In \ parallelogram \ opposite \ angles}$ $\sf{are \ congruent.}$ $\sf{\therefore{\angle DAB=\angle BCD=2x,}}$ $\sf{\angle CBA=\angle CDA}$ $\sf{In \ parallelogram \ sum \ of \ adjacent}$ $\sf{angles \ is \ 180^\circ}$ $\sf{\therefore{\angle DAB+\angle CBA=180^\circ}}$ $\sf{\therefore{2x+2y=180^\circ}}$ $\boxed{\sf{\therefore{x+y=90^\circ}}}$ $\textsf{In parallelogram ABCD,}$ $\sf{Seg \ AJ \ bisects \ \angle DAB,}$ $\sf{Seg \ DI \ bisects \ \angle CDA,}$ $\sf{Seg \ CL \ bisects \ \angle BCD,}$ $\sf{Seg \ BK \ bisects \ \angle CBA}$ $\sf{Now,}$ $\sf{In \ \triangle AID,}$ $\sf{\angle IDA=x}$ $\sf{\angle ADI=y,}$ $\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$ $\sf{\therefore{\angle IDA+\angle ADI+\angle AID=180^\circ}}$ $\sf{\therefore{x+y+\angle AID=180^\circ}}$ $\sf{But, \ x+y=90^\circ}$ $\sf{\therefore{90^\circ+\angle AID=180^\circ}}$ $\sf{\therefore{\angle AID=90^\circ}}$ $\sf{By \ linear \ pair \ axiom}$ $\sf{\angle AID+\angle LIJ=180^\circ}$ $\sf{\therefore{90^\circ+\angle LIJ=180^\circ}}$ $\boxed{\sf{\therefore{\angle LIJ=90^\circ...(1)}}}$ $\sf{In \ \triangle BJA,}$ $\sf{\angle JBA=y,}$ $\sf{\angle JAB=x}$ $\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$ $\sf{x+y+\angle BJA=180^\circ}$ $\sf{\therefore{BJA=90^\circ}}$ $\sf{By \ linear \ pair \ axiom }$ $\boxed{\sf{\angle IJK=90^\circ...(2)}}$ $\sf{In \ \triangle BKC,}$ $\sf{\angle KBC=y,}$ $\sf{\angle KCB=x}$ $\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$ $\sf{x+y+\angle BKC=180^\circ}$ $\sf{\therefore{\angle BKC=90^\circ}}$ $\sf{By \ linear \ pair \ axiom}$ $\boxed{\sf{\angle JKL=90^\circ...(3)}}$ $\sf{In \ \triangle CLD,}$ $\sf{\angle LCD=x,}$ $\sf{\angle LDC=y}$ $\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$ $\sf{x+y+\angle CLD=180^\circ}$ $\sf{\therefore{\angle CLD=90^\circ}}$ $\sf{By \ linear \ pair \ axiom}$ $\boxed{\sf{\angle \ LKJ=90^\circ...(4)}}$ $\sf{In \ \square IJKL,}$ $\sf{From \ (1), \ (2), \ (3) \ and \ (4) \ we \ get}$ $\sf{\angle LIJ=90^\circ,}$ $\sf{\angle IJK=90^\circ,}$ $\sf{\angle JKL=90^\circ,}$ $\sf{\angle LKJ=90^\circ}$ $\sf{Since, all \ angles \ of \ \square IJKL}$ $\sf{are \ of \ 90^\circ \ we \ can \ say}$ $\sf{\square IJKL \ is \ a \ rectangle.}$ $\sf{Hence \ proved,}$ $\sf\purple{\tt{Bisectors \ of \ angles \ of \ the}}$ $\sf\purple{\tt{parallelogram \ form \ a \ rectangle.}}$

Show that bisectors of angles of the parallelogram forms a rectangleplease answer

Answer»

ANSWER:

Solution:

$\sf{Let \ \angle DAB \ be \ 2x \ and \ \angle CBA}$

$\sf{be \ 2y}$

$\sf{In \ parallelogram \ opposite \ angles}$

$\sf{are \ congruent.}$

$\sf{\therefore{\angle DAB=\angle BCD=2x,}}$

$\sf{\angle CBA=\angle CDA}$

$\sf{In \ parallelogram \ sum \ of \ adjacent}$

$\sf{angles \ is \ 180^\circ}$

$\sf{\therefore{\angle DAB+\angle CBA=180^\circ}}$

$\sf{\therefore{2x+2y=180^\circ}}$

$\boxed{\sf{\therefore{x+y=90^\circ}}}$

$\textsf{In parallelogram ABCD,}$

$\sf{Seg \ AJ \ bisects \ \angle DAB,}$

$\sf{Seg \ DI \ bisects \ \angle CDA,}$

$\sf{Seg \ CL \ bisects \ \angle BCD,}$

$\sf{Seg \ BK \ bisects \ \angle CBA}$

$\sf{Now,}$

$\sf{In \ \triangle AID,}$

$\sf{\angle IDA=x}$

$\sf{\angle ADI=y,}$

$\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$

$\sf{\therefore{\angle IDA+\angle ADI+\angle AID=180^\circ}}$

$\sf{\therefore{x+y+\angle AID=180^\circ}}$

$\sf{But, \ x+y=90^\circ}$

$\sf{\therefore{90^\circ+\angle AID=180^\circ}}$

$\sf{\therefore{\angle AID=90^\circ}}$

$\sf{By \ linear \ pair \ axiom}$

$\sf{\angle AID+\angle LIJ=180^\circ}$

$\sf{\therefore{90^\circ+\angle LIJ=180^\circ}}$

$\boxed{\sf{\therefore{\angle LIJ=90^\circ...(1)}}}$

$\sf{In \ \triangle BJA,}$

$\sf{\angle JBA=y,}$

$\sf{\angle JAB=x}$

$\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$

$\sf{x+y+\angle BJA=180^\circ}$

$\sf{\therefore{BJA=90^\circ}}$

$\sf{By \ linear \ pair \ axiom }$

$\boxed{\sf{\angle IJK=90^\circ...(2)}}$

$\sf{In \ \triangle BKC,}$

$\sf{\angle KBC=y,}$

$\sf{\angle KCB=x}$

$\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$

$\sf{x+y+\angle BKC=180^\circ}$

$\sf{\therefore{\angle BKC=90^\circ}}$

$\sf{By \ linear \ pair \ axiom}$

$\boxed{\sf{\angle JKL=90^\circ...(3)}}$

$\sf{In \ \triangle CLD,}$

$\sf{\angle LCD=x,}$

$\sf{\angle LDC=y}$

$\sf{Sum \ of \ all \ angles \ of \ triangle=180^\circ}$

$\sf{x+y+\angle CLD=180^\circ}$

$\sf{\therefore{\angle CLD=90^\circ}}$

$\sf{By \ linear \ pair \ axiom}$

$\boxed{\sf{\angle \ LKJ=90^\circ...(4)}}$

$\sf{In \ \square IJKL,}$

$\sf{From \ (1), \ (2), \ (3) \ and \ (4) \ we \ get}$

$\sf{\angle LIJ=90^\circ,}$

$\sf{\angle IJK=90^\circ,}$

$\sf{\angle JKL=90^\circ,}$

$\sf{\angle LKJ=90^\circ}$

$\sf{Since, all \ angles \ of \ \square IJKL}$

$\sf{are \ of \ 90^\circ \ we \ can \ say}$

$\sf{\square IJKL \ is \ a \ rectangle.}$

$\sf{Hence \ proved,}$

$\sf\purple{\tt{Bisectors \ of \ angles \ of \ the}}$

$\sf\purple{\tt{parallelogram \ form \ a \ rectangle.}}$