Mathematics

In the figure, $$ABCD$$ is a square and $$\triangle EDC$$ is an equilateral triangle. Prove that
(i) $$AE=BE$$
(ii) $$\angle DAE={15}^{o}$$


SOLUTION
(i) From the figure we know that $$\triangle EDC$$ is an equilateral triangle
so we get $$\angle EDC=\angle ECD={60}^{o}$$

We know that $$ABCD$$ is a square

so we get $$\angle CDA=\angle DCB={90}^{o}$$

consider $$\triangle EDA$$

we get

$$\angle EDA={60}^{o}+{90}^{o}$$

so we get

$$\angle EDA={150}^{o}....(1)$$

Consider $$\triangle ECB$$

we get

$$\angle ECB=\angle ECD+\angle DCB$$

by substituting the values in the above equation

$$\angle ECB={60}^{o}+{90}^{o}$$

we get

$$\angle ECB={150}^{o}$$

so we get $$\angle EDA=\angle ECB....(ii)$$

consider $$\triangle EDA$$ and $$\triangle ECB$$

from the figure we know that $$ED$$ and $$EC$$ are the sides of an equilateral triangle

so we get

$$ED=EC$$

we also know that the sides of the square are equal

$$DA=CB$$

by SAS congruence criterion

$$\triangle EDA\cong \triangle ECB$$

$$AE=BE$$

(ii) consider $$\triangle EDA$$

we know that

$$ED=DA$$

from the figure we know that the base angles are equal

$$\angle DEA=\angle DAE$$

based on equation (i) we get $$\angle EDA={150}^{o}$$

by angle sum property

$$\angle EDA+\angle DAE+\angle DEA={180}^{o}$$

by substituting the values we get

$${150}^{o}+\angle DAE+\angle DEA={ 180 }^{ o }$$

we know that $$\angle DEA=\angle DAE$$

so we get

$${150}^{o}+\angle DAE+\angle DAE={ 180 }^{ o }$$

$$2\angle DEA={ 180 }^{ o }-{150}^{o}$$

$$\angle DAE={15}^{o}$$
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