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