In the figure, ABCDEF is a regular hexagon. Prove that triangle BDF, drawn joining alternate
vertices is equilateral
Answers
Answer:
Triangle is equilateral.
Step-by-step explanation:
Construction: Draw a perpendicular AO on BF
All the angles are 120° as it is regular hexagon and all the sides are equal.
Now in Δ AOB and Δ AOF,
AO = AO [common]
AB = AF [all sides are equal]
∠ AOB = ∠ AOF = 90° [by construction]
So, Δ AOB ≅ Δ AOF
∠ BAO = ∠ FAO = 60° [CPCT]
∠ ABO = ∠ AFO [CPCT] … (1)
Similarly, Δ BGC ≅ Δ DGC
∠ BCG = ∠ DCG = 60°
∠ CBG = ∠ CDG … (2)
Similarly, Δ DHE ≅ Δ FHE
∠ DEH = ∠ FEH = 60°
∠ EDH = ∠ EFH … (3)
In Δ BAO,
∠ BAO + ∠ AOB + ∠ ABO = 180°
⇒ 60° + 90° + ∠ ABO = 180°
⇒ 150° + ∠ ABO = 180°
⇒ ∠ ABO = 180° - 150°
⇒ ∠ ABO = 30°
∴ from (1) ∠ ABO = ∠ AFO = 30°
Similarly
∠ CBG = ∠ CDG = 30°
∠ EDH = ∠ EFH = 30°
∠ ABO + ∠ FBD + ∠ CBG = ∠ ABC
⇒ 30° + ∠ FBD + 30° = 120°
⇒ ∠ FBD + 60° = 120°
⇒ ∠ FBD = 120° - 60°
⇒ ∠ FBD = 60°
Similarly,
∠ BDF = DFB = 60°
In Δ BDF
∠ FBD = ∠ BDF = DFB = 60°
∴ Δ BDF is an equilateral triangle.
Step-by-step explanation:
All the angles are 120° as it is regular hexagon and all the sides are equal.
Now in Δ AOB and Δ AOF,
AO = AO [common]
AB = AF [all sides are equal]
∠ AOB = ∠ AOF = 90° [by construction]
So, Δ AOB ≅ Δ AOF
∠ BAO = ∠ FAO = 60° [CPCT]
∠ ABO = ∠ AFO [CPCT] … (1)
Similarly, Δ BGC ≅ Δ DGC
∠ BCG = ∠ DCG = 60°
∠ CBG = ∠ CDG … (2)
Similarly, Δ DHE ≅ Δ FHE
∠ DEH = ∠ FEH = 60°
∠ EDH = ∠ EFH … (3)
In Δ BAO,
∠ BAO + ∠ AOB + ∠ ABO = 180°
⇒ 60° + 90° + ∠ ABO = 180°
⇒ 150° + ∠ ABO = 180°
⇒ ∠ ABO = 180° - 150°
⇒ ∠ ABO = 30°
∴ from (1) ∠ ABO = ∠ AFO = 30°
Similarly
∠ CBG = ∠ CDG = 30°
∠ EDH = ∠ EFH = 30°
∠ ABO + ∠ FBD + ∠ CBG = ∠ ABC
⇒ 30° + ∠ FBD + 30° = 120°
⇒ ∠ FBD + 60° = 120°
⇒ ∠ FBD = 120° - 60°
⇒ ∠ FBD = 60°
Similarly,
∠ BDF = DFB = 60°
In Δ BDF
∠ FBD = ∠ BDF = DFB = 60°
∴ Δ BDF is an equilateral triangle
