Question

14. ABCDE is a regular pentagon The bidector of
angle A of the pentagon meets the side CD in
point M. Show that ZAMC = 90
D
A.​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

so ABCDE is a regular Pentagon

so each angle of pentagon will be = ( 5-2)× 180

= 540 deg

now it has 5 angles

let each equal angle be = x

so x+x+x+x+x = 540

5x = 540

x = 540/5 = 108 deg

and x/2= 108/2 = 54 deg

now bisected angle A = 54 deg

and bisection of angle A will lead to formation of two quardilaterals ABCM and AEDM

so angle AMC in ABCM can be found as

ZAMC + ZC+ZB+ZA/2 = 360 deg ( angle sum property of quardilateral)

now as it is a regular Pentagon

ZC = ZB = 108 deg

ZA/2 = 54

so

ZAMC = 360 - ( 108 + 108 + 54)

= 360 - 270

ZAMC = 90 deg

thank you

see image for better understanding