`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.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^@ .

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

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

The side B C of A B C is produced to a point Ddot The bisector of /_A meets side B C in Ldot If /_A B C=30^0a n d /_A C D=115^0, then /_A L C= 85^0 (b) 72 1/2^0 (c) 145^0 (d) none of these

The side B C of A B C is produced to a point Ddot The bisector of /_A meets side B C in Ldot If /_A B C=30^0a n d\ /_A C D=115^0, then /_A L C= 85^0 (b) 72 1/2^0 (c) 145^0 (d) none of these

The sides BC of Delta A B C is produced to a point D . The bisector of /_A meets side B C in L . If /_A B C=30^0\ a n d\ /_A C D=115^0, find /_A L C

In a parallelogram A B C D , the bisectors of /_A\ a n d\ /_B meet at Odot Find /_A O Bdot

A D is a median of Delta A B C . The bisector of /_A D B and /_A D C meet AB and AC in E and F respectively. Prove that EF||BC

In a triangle A B C ,\ /_A B C=\ /_A C B and the bisectors of /_A B C\ a n d\ /_A C B intersect at O such that /_B O C=120^0dot Show that /_A=/_B=/_C=60^0 .