Question

In the given figure (not drawn to scale), ABCD is a square such that AE=DE.Find $\angle BEC.$

• A. ${28}^0$
• B. ${56}^0$
• C. ${62}^0$
• D. ${24}^0$

Answer 1

The correct option is B ${56}^0$

from square ABCD , $AE=ED$

hence $BE=EC$ hence $<EBC=<ECB$ ( two sides of triangle is equal then opposite two angles are equal.)

now$<EBC=({90}^0-<ABE)=({90}^0-{28}^0)={62}^0$

$<EBC=<ECB={62}^0$

now $<BEC={180}^0-(<EBC+<ECB)={180}^0-{62}^0-{62}^0={56}^0$

hence option B is correct.