ABCDE is a pentagon in which AB = AE, BC = ED and ∠ABC = ∠AED.

(a) Prove that :

(i) AC = AD

(ii) ∠BCD = ∠EDC

(b) If BC and ED are produced to meet at X, prove that BX = EX.

(a) Given: ABCDE is a pentagon in which AB = AE, BC = ED and ∠ABC = ∠AED.

To prove: (i) AC = AD

(ii) ∠BCD = ∠EDC

Proof: Consider Δ AED and Δ ABC,

AE = AB (Given)

ED = BC (Given)

∠AED = ∠ABC (Given)

Using SAS congruency criterion,

Δ AED ≅ Δ ABC

⇒ AD = AC (by C.P.C.T.)

Hence, AD = AC.

(ii) Since Δ AED ≅ Δ ABC (Proved above), we have:

⇒ ∠ADE = ∠ACB (by C.P.C.T.) ……………….(1)

Since AD = AC (proved above), Δ ACD is isosceles, so:

⇒ ∠ADC = ∠ACD ……………….(2)

Adding equations (1) and (2), we get:

⇒ ∠ADC + ∠ADE = ∠ACD + ∠ACB

⇒ ∠EDC = ∠BCD

Hence, ∠EDC = ∠BCD.

(b)

Since BC and ED are produced to meet at X, consider Δ XDC:

⇒ ∠XDC = 180° - ∠EDC

⇒ ∠XDC = 180° - ∠BCD (∠EDC = ∠BCD)

⇒ ∠XDC = ∠XCD

⇒ XC = XD ……………….(3)

As it is given , BC = DE ……………….(4)

Adding equations (3) and (4),

⇒ XC + BC = XD + DE

⇒ BX = EX

Hence, BX = EX.