Question

Abcde is a regular pentagon.The bisector of anglea of the pentagon meets the side cd in m.Show that angle amc=90 degree

Answer 1

Answer:

As per Question : ABCDE is a regular pentagon. The bisector ∠A of the pentagon meets the side CD at point M.

To prove : ∠AMC = 90°

Proof: We know - measure of each interior angle of a regular pentagon is 108°.

∠BAM = 1/2 X 108° = 54°

Since, the sum of a quadrilateral is 360°

In quadrilateral ABCM,

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

54° + 108° + 108° + ∠AMC = 360°

∠AMC = 360° – 270°

∠AMC = 90°