91Ó°ÊÓ

The given diagram is:

It is being given that $ABCD,DEFG,CNME$are the squares and$CDE$is an equilateral triangle.

As, the triangle$CDE$ is an equilateral triangle, therefore all the sides of the triangle$CDE$ are equal.

Therefore, $CD=DE=EF$.

As,$ABCD$ is a square, therefore $AD=CD$.

As,$DEFG$is a square, therefore $DE=EF$.

As, $CNME$is a square, therefore $EC=CN$.

As,$CD=DE=EF,AD=CD,DE=EF$ and $EC=CN$, therefore it can be obtained that $AD=CD=DE=EF=EC=CN$.

Now, in the triangle $\mathrm{Δ}ADE$, it can be noticed that $AD=DE$, therefore the angles opposite to the equal sides must be equal.

Therefore, it can be noticed that $∠DAEâ‰\dots ∠DEA$.

In the triangle $\mathrm{Δ}FEC$, it can be noticed that $EF=EC$, therefore the angles opposite to the equal sides must be equal.

Therefore, it can be noticed that $∠ECFâ‰\dots ∠EFC$.

In the triangle $\mathrm{Δ}DCN$, it can be noticed that $DC=CN$, therefore the angles opposite to the equal sides must be equal.

Therefore, it can be noticed that $∠CNDâ‰\dots ∠CDN$.

Now it can be noticed that in the triangles $\mathrm{Δ}ADE$and $\mathrm{Δ}FEC$,

Now as $AD=CD=DE=EF=EC=CN,∠DAEâ‰\dots ∠DEA,∠ECFâ‰\dots ∠EFC$and $∠CNDâ‰\dots ∠CDN$, therefore it can be noticed that $∠DAEâ‰\dots ∠DEAâ‰\dots ∠ECFâ‰\dots ∠EFCâ‰\dots ∠CNDâ‰\dots ∠CDN$.

Therefore, in the triangles $\mathrm{Δ}ADE$and width="46" height="19" role="math">ΔFEC, it can be noticed that:

$ADâ‰\dots FE,∠DAEâ‰\dots ∠EFC$and $∠DEAâ‰\dots ∠ECF$.

Therefore the triangles$\mathrm{Δ}ADE$and width="46" height="19" role="math">ΔFEC are the congruent triangles by using the AAS postulate.

Similarly, the triangles$\mathrm{Δ}ADE$ and$\mathrm{Δ}DCN$ are the congruent triangles by using the AAS postulate.

As,width="109" height="20" role="math">ΔADE≅ΔFECand $\mathrm{Δ}ADEâ‰\dots \mathrm{Δ}DCN$, therefore $\mathrm{Δ}ADEâ‰\dots \mathrm{Δ}FECâ‰\dots \mathrm{Δ}DCN$.

Therefore by using the corresponding parts of the congruent triangles it can be said that $\stackrel{¯}{AE}â‰\dots \stackrel{¯}{FC}â‰\dots \stackrel{¯}{ND}$.

The two-column proof is: