in a regular pentagon ABCDE , draw a diagonal BE and then find thenmeasures of :angle BAE , angle ABE and angle BED answer needed asnfast as possible
Given: ABCDE is a regular pentagon
⇒ Sum of angles of ABCDE = 180° × (5 – 2) = 540°
⇒ Each angle of ABCDE =
⇒ ∠BAE = 108°
Now in ΔABE
AB = AE
⇒ ∠ABE = ∠AEB
and ∠BAE + ∠ABE + ∠AEB = 180°
⇒ 108° + ∠ABE + ∠ABE = 180°
⇒ 2∠ABE = 180° – 108°
and ∠AED = ∠AEB + ∠BED
⇒ ∠BED = ∠AED – ∠AEB
= 108° – 36°
= 72°