Question

ABCDE is a regular polygon , The bisector of angle A meets CD at point M . Prove that angle AMC is 90 degres

Answer 1

Step-by-step explanation:

we know that

angle sum of regular pentagon is 540

therefore each angle 540/5=108

AEDM is a quadrilateral

their angle sum is 360

we know ,

angle d =108

angle E=108

angle EAD= 108/2

=54

let angle m be x degree

their sum is 360

that is d + E + EAD +M=360

108+108+54+x=360

270+x=360

x=360-270

x =90

therfore bisector of angle A intersect CD at M with 90 degrees

Answer 2

Answer:

See, We have circumscribed the regular pentagon with a circle.

Since, the 360 of the circle is divided equally by the pentagon to five part i.e.,72 the angle COD =72.

Now, We have calculated angle CAD=36,

which is half of angle COD, using the property of a circle

Now, We have calculated angle ACD, using property of triangle, in triangle ACD.

36+angle ACD+angle ADC=180

angleACD=70.

Now finally take a look at the triangle AMC

Angle MAC=half of angle CAD=18

Now,

angle MAC+angleAMC+angleACM=180

18+angle AMC+72=180

angle AMC=90

Hence proved.

Answer 3

Answer:

See, We have circumscribed the regular pentagon with a circle.

Since, the 360 of the circle is divided equally by the pentagon to five part i.e.,72 the angle COD =72.

Now, We have calculated angle CAD=36,

which is half of angle COD, using the property of a circle

Now, We have calculated angle ACD, using property of triangle, in triangle ACD.

36+angle ACD+angle ADC=180

angleACD=70.

Now finally take a look at the triangle AMC

Angle MAC=half of angle CAD=18

Now,

angle MAC+angleAMC+angleACM=180

18+angle AMC+72=180

angle AMC=90

Hence proved.