Question

show that bisectors of angles of the parallelogram forms a rectangle
please answer ​

Answer 1

Answer:

Solution:

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

$\sf{be \ 2y}$

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

$\sf{are \ congruent.}$

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

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

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

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

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

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

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

$\textsf{In parallelogram ABCD,}$

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

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

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

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

$\sf{Now,}$

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

$\sf{\angle IDA=x}$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sf{\angle JAB=x}$

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

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

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

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

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

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

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

$\sf{\angle KCB=x}$

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

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

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

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

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

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

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

$\sf{\angle LDC=y}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sf{Hence \ proved,}$

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

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

Answer 2

Step-by-step explanation:

Answer:

Solution:

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

\sf{be \ 2y}be 2y

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

\sf{are \ congruent.}are congruent.

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

\sf{\angle CBA=\angle CDA}∠CBA=∠CDA

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

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

∘

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

∘

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

∘

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

∴x+y=90

∘

\textsf{In parallelogram ABCD,}In parallelogram ABCD,

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

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

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

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

\sf{Now,}Now,

\sf{In \ \triangle AID,}In △AID,

\sf{\angle IDA=x}∠IDA=x

\sf{\angle ADI=y,}∠ADI=y,

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

∘

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

∘

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

∘

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

∘

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

∘

+∠AID=180

∘

\sf{\therefore{\angle AID=90^\circ}}∴∠AID=90

∘

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

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

∘

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

∘

+∠LIJ=180

∘

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

∴∠LIJ=90

∘

...(1)

\sf{In \ \triangle BJA,}In △BJA,

\sf{\angle JBA=y,}∠JBA=y,

\sf{\angle JAB=x}∠JAB=x

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

∘

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

∘

\sf{\therefore{BJA=90^\circ}}∴BJA=90

∘

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

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

∠IJK=90

∘

...(2)

\sf{In \ \triangle BKC,}In △BKC,

\sf{\angle KBC=y,}∠KBC=y,

\sf{\angle KCB=x}∠KCB=x

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

∘

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

∘

\sf{\therefore{\angle BKC=90^\circ}}∴∠BKC=90

∘

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

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

∠JKL=90

∘

...(3)

\sf{In \ \triangle CLD,}In △CLD,

\sf{\angle LCD=x,}∠LCD=x,

\sf{\angle LDC=y}∠LDC=y

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

∘

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

∘

\sf{\therefore{\angle CLD=90^\circ}}∴∠CLD=90

∘

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

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

∠ LKJ=90

∘

...(4)

\sf{In \ \square IJKL,}In □IJKL,

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

\sf{\angle LIJ=90^\circ,}∠LIJ=90

∘

,

\sf{\angle IJK=90^\circ,}∠IJK=90

∘

,

\sf{\angle JKL=90^\circ,}∠JKL=90

∘

,

\sf{\angle LKJ=90^\circ}∠LKJ=90

∘

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

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

∘

we can say

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

\sf{Hence \ proved,}Hence proved,

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