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Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC at F.

Construction: Join EC and AD.

Proof: ∵ ΔABC is an equilateral triangle. ∴ ∠ABC = ∠BCA = ∠CAB = 60° ...(1)

∵ ΔBDE is an equilateral triangle.

∴ ∠BDE = ∠DEB = ∠EBD = 60° ...(2) ∠ABE + ∠BED

= ∠ABD + ∠EBD + ∠BED = 60°+ 60°+ 60°= 180°

∴ AB || DE ...(3)

∵ Sum of consecutive interior angles on

the same side of a transversal is 180° ∠EBA + ∠BAC

= ∠EBD + ∠DBA + ∠BAC = 60° + 60° + 60° = 180°

∴ AC || BE ...(4)

∵ Sum of consecutive interior angles on the same side of the transversal is 180°

∵ ΔCBA and ΔCEA are on the same base AC and between the same parallels.

∴ ar(ΔCBA) = ar(ΔCEA)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area ⇒ ar(ΔABC) = ar(ΔCDA) + ar(ΔCED) + ar(ΔADE) ...(5)

In ΔABC,

∵ AD is a median.

∵ A median of a triangle divides it into two triangles of equal area

In ΔEBC,

∵ ED is a median.

∵ A median of a triangle divides it into two triangles of equal area ∵ ΔDEA and ΔDBE are on the same base DE and between the same parallels AB and DE.

∴ ar(ΔDEA) = ar(ΔDBE) ...(8)

∵ Two triangles on the same base (or equal bases) and between the same parallels are equal in area

Using (6), (7) and (8), (5) gives

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