Question

4. In the adjoining figure, ABCD is a square and AEDC is an equilateral triangle. Prove that (i) AE = BE, (i) DAE = 15​

Answer 1

Step-by-step explanation:

We know that ABCD is a square

so we get ∠CDA=∠DCB=90

o

consider △EDA

we get

∠EDA=60

o

+90

o

so we get

∠EDA=150

o

....(1)

Consider △ECB

we get

∠ECB=∠ECD+∠DCB

by substituting the values in the above equation

∠ECB=60

o

+90

o

we get

∠ECB=150

o

so we get ∠EDA=∠ECB....(ii)

consider △EDA and △ECB

ED=EC (sides of an equilateral triangle)

∠EDA=∠ECB (From (ii))

DA=CB (sides of square)

by SAS congruence criterion

△EDA≅△ECB

AE=BE

(ii) consider △EDA

we know that

ED=DA

from the figure we know that the base angles are equal

∠DEA=∠DAE

based on equation (i) we get ∠EDA=150

o

by angle sum property

∠EDA+∠DAE+∠DEA=180

o

by substituting the values we get

150

o

+∠DAE+∠DEA=180

o

we know that ∠DEA=∠DAE

so we get

150

o

+∠DAE+∠DAE=180

o

2∠DEA=180

o

−150

o

∠DAE=15

o