` ABCDE` is a regular pentagon. The bisector of angle `A` meets side `CD` at `P`, then prove that `APC = 90^@.`

ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find angle AMC

ABCDE is a regular pentagon.The bisector of /_A of the pentagon meets the side CD in M* Show that /_AMC=90^(@).

ABCDE is a regular pentagon. The bisector of angle A of the pentagon meet the side CD in point M. Show that angleAMC=90^@ .

A B C D E is a regular pentagon. The bisector of /_A of the pentagon meets the side C D in Mdot Show that /_A M C=90^0dot

ABCDE is a regular pentagon. Angle bisector of angle BAE meets CD at m. angle bisector of angle BCD meets AM at P. Find the angle CPM = ?

ABCDE is a regular pentagon and bisector of BAE meets CD at M, bisector of BCD meets AM at P. Prove that CPM=36^(@).

In parallelogram ABCD, the bisector of angle A meets DC at P and AB= 2AD. Prove that: Angle APB= 90^(@)

ABCDE is a regular pentagon and bisector of /_BAE meets CD at M. If bisector of /_BCD meets AM at P, find /_CPM.

A B C D E is a regular pentagon and bisector of /_B A E\ meets C D\ at Mdot If bisector of /_B C D meets A M at P , find /_C P Mdot

Angle of rotation in a regular pentagon is .