Question

please can anyone solve this maths question ... i will​ definitely mark your ans as brainelist...thankyou... your lovingly friend- Tanya..
1) prove that :- AE = BE.
2) Angle DAE = 15°

Answer 1

one side of the Δdec is joined with the square abcd 
∴AB=BC=CD=DA=DE=EC
∴in ΔEDA and ΔECB 
ED = EC , DA= BC 
∠EDC = 60° ∠ADC = 90° ∴∠EDA = 150°
similarly ∠ECB = 150°
∴ΔEDC = ΔECB 
∴AE = BE ....... (cpct)
in ΔADE 
∠ADE +∠DAE +∠DEA = 180°
150° + 2∠DAE = 180°   ........ ( ΔADE IS an isoseles Δ)
2∠DAE = 30°
∴∠DAE = 15°

Answer 2

May this will help you