Question

In the given figure, \[ABCD\] is a square and \[\Delta DEC\] is an equilateral triangle. Prove that:
(i) \[\Delta ADE\] is congruent to \[\Delta BCE\]
(ii) \[AE = BE\]
(iii) \[\angle DAE = 15^\circ \]

Answer 1

Hint:
 Here, we will use the properties of a square and equilateral triangle to prove that \[\Delta ADE\] is congruent to \[\Delta BCE\]. Using the congruent parts of congruent triangles, we will prove that \[AE = BE\]. Since one of the sides of the square and equilateral triangle is the same, all sides of the square and the equilateral triangle are equal to each other. We will use the isosceles triangle property and the angle sum property of a triangle to prove that \[\angle DAE = 15^\circ \]. According to the isosceles triangle property, the angles opposite to equal sides of a triangle are equal. According to the angle sum property of a triangle, the sum of all the interior angles of a triangle is always \[180^\circ \].

Complete step by step solution:
(i)
First, we will use the properties of a square and equilateral triangle to prove that \[\Delta ADE\] is congruent to \[\Delta BCE\].
We know that all sides of a square are equal, and all interior angles of a square measure \[90^\circ \].
Therefore, since \[ABCD\] is a square, we get
\[AB = BC = CD = DA\]
\[\angle ABC = \angle BCD = \angle CDA = \angle DAB = 90^\circ \]
We know that all sides of an equilateral triangle are equal, and all interior angles of an equilateral triangle measure \[60^\circ \].
Therefore, since \[\Delta DEC\] is an equilateral triangle, we get
\[DE = EC = CD\]
\[\angle DEC = \angle ECD = \angle CDE = 60^\circ \]
Now, we can observe that \[\angle BCD = \angle CDA\] and \[\angle ECD = \angle CDE\].
We know that equals added to equals remain equal.
Therefore, we get
\[\angle BCD + \angle ECD = \angle CDA + \angle CDE\]
Using the figure, we can rewrite this equation as
\[\angle BCE = \angle ADE\]
Now, we will prove that the triangles \[\Delta ADE\] and \[\Delta BCE\] are congruent.
In \[\Delta ADE\] and \[\Delta BCE\], we have
\[AD = BC\] (Sides of the same square)
\[DE = CE\] (Sides of the same equilateral triangle)
\[\angle BCE = \angle ADE\]
\[\therefore \Delta ADE\cong \Delta BCE\] (by the S.A.S. congruence criterion)
Therefore, by S.A.S. congruence criterion, we have proved that triangles \[\Delta ADE\] and \[\Delta BCE\] are congruent.
(ii)
Now, we will prove that \[AE = BE\].
We know that congruent parts of two congruent triangles are equal.
Since \[\Delta ADE\] and \[\Delta BCE\] are congruent, we get
\[\therefore AE=BE\]
Therefore, we have proved that \[AE = BE\].
(iii)
We know that \[AB = BC = CD = DA\] and \[DE = EC = CD\].
Since \[CD\] is there in both, we get
\[AB = BC = CD = DA = DE = EC\]
Now, in triangle \[\Delta ADE\], we have
\[DA = DE\]
We know that the angles opposite to equal sides of a triangle are equal.
Therefore, we get
\[\angle DAE = \angle DEA\]
Finally, we will apply the angle sum property of a triangle in \[\Delta ADE\].
According to the angle sum property of a triangle, the sum of all the interior angles of a triangle is always \[180^\circ \].
Therefore, we get
\[\angle DAE + \angle DEA + \angle ADE = 180^\circ \]
Rewrite \[\angle ADE\] as \[\angle ADC + \angle EDC\], we get
\[ \Rightarrow \angle DAE + \angle DEA + \angle ADC + \angle EDC = 180^\circ \]
Substituting \[\angle DAE = \angle DEA\], \[\angle ADC = 90^\circ \], and \[\angle EDC = 60^\circ \], we get
\[ \Rightarrow \angle DAE + \angle DAE + 90^\circ + 60^\circ = 180^\circ \]
Adding the like terms, we get
\[ \Rightarrow 2\angle DAE + 150^\circ = 180^\circ \]
Subtracting \[150^\circ \] from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow 2\angle DAE + 150^\circ - 150^\circ = 180^\circ - 150^\circ \\ \Rightarrow 2\angle DAE = 30^\circ \end{array}\]
Dividing both sides by 2, we get
\[\therefore \angle DAE=15{}^\circ \]

Hence, we have proved that \[\Delta ADE \cong \Delta BCE\], \[AE = BE\], and \[\angle DAE = 15^\circ \].

Note:
We used the S.A.S. congruence criterion to prove that \[\Delta ADE \cong \Delta BCE\]. According to the S.A.S. congruence criterion, if in two triangles, any two corresponding sides and the angle formed by them at the common vertex are equal, then the two triangles are congruent.