Question

$ABCD$ is a square and $△DEC$ is an equilateral triangle. $△ADE\cong △BCE$ by

• A. $RHS$
• B. $SSS$
• C. $AAS$
• D. $SAS$

Answer 1

Answer: (D)

Given,
$\angle ADC=\angle BCD={90}^{\circ }$ ...... (Angles of a square)
Also, $\angle EDC=\angle ECD={60}^{\circ }$ ...... (Angles of equilateral triangle)
Adding the two equations,
$\angle ADC+\angle EDC=\angle BCD+\angle ECD$
$\angle D=\angle C$ (1)
Now, in $△AED$ and $△BEC$,
$AD=BC$ (Sides of a square)
$ED=EC$ (Sides of an equilateral triangle)
$\angle D=\angle C$ (From 1)
Thus, $△AED\cong △BEC$ ...... (By SAS rule)