Question

in the figure ABCDE is a regular pentagon bisectors of anlgle A meets CD at M prove that angle AMC = 90 degree

Answer 1

we know that sum of interior and exterior of a pentagon is 540° [(n-2)×180] and each angle = 108° .

therefore ..bisector angle a =54°
now in quadrilateral abcm,
a+b+c+m=180°
therefore
54+108+108+amc=360
270+amc=360
amc= 360-270=90
hence proved!.

Answer 2

Answer:

Step-by-step explanation:

we know that sum of interior and exterior of a pentagon is 540° [(n-2)×180] and each angle = 108° .

therefore ..bisector angle a =54°

now in quadrilateral abcm,

a+b+c+m=180°

therefore

54+108+108+amc=360

270+amc=360

amc= 360-270=90

hence proved!!!!

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