Question

in the adjoining figure ABCD is a square and Triangle EDC is an equilateral triangle prove that AE=BE, angle DAE=15°​

Answer 1

Answer:

(i) From the figure we know that AEDC is an equilateral triangle so we get ZEDC = ZECD = 60° We know that ABCD is a square so we get CDA = ZDCB = 90° consider AEDA Feedback we get ZEDA = 60° + 90⁰ so we get ZEDA= 150°....(1) Consider AECB

Consider AECB we get ZECB: = ZECD + 2DCB by substituting the values in the above equation ZECB = 60° + 90° we get ZECB = 150⁰ so we get ZEDA = ZECB....(ii) consider AEDA and AECB ED = EC (sides of an equilateral triangle) /EDA = /ECB (From (ii)

ED = EC (sides of an equilateral triangle) ZEDA= ZECB (From (ii)) DA = CB (sides of square) by SAS congruence criterion AEDA AECB AE = BE (ii) consider AEDA we know that ED = DA from the figure we know that the base angles are equal ZDEA = ZDAE

ZDEA = DAE based on equation (i) we get ZEDA= 150⁰ by angle sum property ZEDA+ZDAE + <DEA = 180° by substituting the values we get 150° + ZDAE + ZDEA = 180° we know that ZDEA = ZDAE so we get 150° + ZDAE + ZDAE = 180° 2ZDEA = 180° - 150⁰

150⁰ + DAE + ZDAE = 180° 2ZDEA = 180° - 150° ZDAE = 15°