Question

in the given figure BM bisects angle B and AN bisects angle A of parallelogram ABCD.prove that ABNM is rhombus​

Answer 1

Step-by-step explanation:

Given parallelogram ABCD in which diaginal BD bisects ∠B and ∠D

BD bisect ∠B,

⇒∠ABD=∠CBD=

2

1

∠ABC

BD bisect ∠D,

⇒∠ADB=∠CDB=

2

1

∠ADC

∴∠ABC=∠ADC [ Opposite angles of parallelogram are equal ]

∴

2

1

∠ABC=

2

1

∠ADC

⇒∠ABD=∠ADB and ∠CBD=∠CDB

∴AD=AB and CD=BC

As ABCD is a parallelogram, opposite sides are equal

⇒AB=CD and AD=BC

⇒AB=BC=CD=AD i.e, ABCD is a rhombus.

Hence, the answer is proved.