Question

In the figure ABCD is a square ∆CDE & ∆ BDF are two equilateral triangles. Prove that
area (A CDE) => 1/2
area (A BDF).​

Answer 1

Answer:

According to figure,

∠ADC=∠BCD=90

∘

→square

∠CDE=∠DCE=60

∘

→equilateral_triangle

∴∠ADE=∠BCE=150

∘

InΔADEandΔBCE

AD=BC

DE=CE

∠ADE=∠BCE

ΔADE≅ΔBCE

Answer 2

Answer:

According to figure,

∠ADC=∠BCD=90

∘

→square

∠CDE=∠DCE=60

∘

→equilateral_triangle

∴∠ADE=∠BCE=150

∘

InΔADEandΔBCE

AD=BC

DE=CE

∠ADE=∠BCE

ΔADE≅ΔBCE