Question

ABCD is a square and DEC is an equilateral triangle. prove that angle AED =15°

Answer 1

hey mate
here is your answer

$<b><i> as ABCD is a square and DCE is a equilateral triangle</b>$

there all sides will be equal.
square ABCD and ∆DEC share side CD
so, side of square= side of triangle

AD=DE
angle of a square=90°
angle of a equilateral ∆=60°

angle ADE=90+60=150°

in ∆ADE
AD=DE .......proved
angle dae=angle dea

angle dae + angle dea +angle ade=180°
2angle dae =30°

angle DAE = 15° =angle AED

hope it helps u :) ☺☺☺
plzzz mark as brainliest ✌✌✌✌❤❤❤

Answer 2

hey mate

here is your answer

as ABCD is a square and DCE is a equilateral triangle

there all sides will be equal.

square ABCD and ∆DEC share side CD

so, side of square= side of triangle

AD=DE

angle of a square=90°

angle of a equilateral ∆=60°

angle ADE=90+60=150°

in ∆ADE

AD=DE .......proved

angle dae=angle dea

angle dae + angle dea +angle ade=180°

2angle dae =30°

angle DAE = 15° =angle AED

hope it helps u :) ☺☺☺

plzzz mark as brainliest ✌✌✌✌❤❤❤