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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Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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