Question

ABCDE is a regular pentagon in which AC and EB intersect at P. Find angle EAP and angle BPA. Also prove that EA = EP.

50 points.
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Answer 1

Answer:

ABCDE is a regular pentagon ⇒ AB = BC = CD = DE = EA, ∠A = ∠B = ∠C = ∠D = ∠E = 108° In ∆ABE, AE = AB ⇒ ∠AEB = ∠ABE [Angles opposite to equal sides are equal.] ------- (1) In ∆ABC, AB = BC ⇒ ∠BAC = ∠BCA [Angles opposite to equal sides are equal.] -----(2) Consider ∆ABE, ∠AEB + ∠ABE + ∠EAB = 180° [Angle sum property of the triangle] ∠AEB + ∠AEB + 108° = 180° [From equation 1] 2∠AEB = 180° − 108° 2∠AEB = 72° ∠AEB = 72° / 2 ∠AEB = 36° ------- (3) Consider ∆ABC ∠BAC + ∠BCA + ∠ABC = 180° [Angle sum property] ∠BAC + ∠BAC + 108° = 180° [From equation 2] 2∠BAC = 180° − 108° 2∠BAC = 72° ∠BAC = 72°/2 ∠BAC = 36° ------- (4) ∠EAP + ∠BAP = 108° ∠EAP + 36° = 108° ∠EAP = 108° − 36° ∠EAP = 72° ------- (5) Thus, ∠EAP = 72° Consider ∆APB: ∠PAB + ∠ABP + ∠BPA = 180° [Angle sum property] 36° + 36° + ∠BPA = 180° 72° + ∠BPA = 180° ∠BPA = 180° − 72° ∠BPA = 108° Thus, ∠BPA = 108° Similarly, In ∆EAP ∠AEP + ∠EAP + ∠EPA = 180° 36° + 72° + ∠EPA = 180° 108° + ∠EPA = 180° ∠EPA = 180° − 108° ∠EPA = 72° ∠EPA = ∠EAP = 72° ⇒ EA = EP [Sides opposite to equal angles are equal] Therefore, EA = EP

Answer 2

angle EAP and angle BPA are ultranative angle therefore they both are equal

<EAP=<BPA

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