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 $DE=FG$.

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$\mathrm{Δ}FEC$ 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,$\mathrm{Δ}ADEâ‰\dots \mathrm{Δ}FEC$ and $\mathrm{Δ}ADEâ‰\dots \mathrm{Δ}DCN$, 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 $\stackrel{¯}{AE}â‰\dots \stackrel{¯}{FC}â‰\dots \stackrel{¯}{ND}$.

From the given diagram it can be noticed that:

$∠AEF=∠AED+∠DEF$

As, $DEFG$is a square, therefore $∠DEF=90°$.

Therefore, it can be obtained that:

$$\begin{gathered} ∠AEF=∠AED+∠DEF \\ =∠AED+90° \end{gathered}$$

From the given diagram it can be noticed that:

$∠NDA=∠NDC+∠CDA$

As, $ABCD$is a square, therefore $∠CDA=90°$.

Therefore, it can be obtained that:

$$\begin{gathered} ∠NDA=∠NDC+∠CDA \\ =∠NDC+90° \end{gathered}$$

As, $∠AEDâ‰\dots ∠NDC$, therefore it can be obtained that:

$$\begin{gathered} ∠AEF=∠AED+90° \\ =∠NCD+90° \end{gathered}$$

Therefore, $∠AEF=∠NDA$.

In the triangles $\mathrm{Δ}AEF$and $\mathrm{Δ}NDA$, it can be noticed that:

$AEâ‰\dots ND,∠AEFâ‰\dots ∠NDA$and$EFâ‰\dots DA$

Therefore, the triangles $\mathrm{Δ}AEF$and $\mathrm{Δ}NDA$are the congruent triangles by using the SAS postulate.

Therefore by using the corresponding parts of congruent triangles, it can be said that: $AFâ‰\dots NA.$

From the given diagram it can be noticed that:

$∠FCN=∠FCE+∠ECN$

As,$CNME$is a square , therefore $∠ECN=90°$.

Therefore, it can be obtained that:

role="math" localid="1649945044141" $$\begin{gathered} ∠FCN=∠FCE+∠ECN \\ =∠FCE+90° \end{gathered}$$

From the given diagram it can be noticed that:

$∠NDA=∠NDC+∠CDA$

As,$ABCD$ is a square , therefore $∠CDA=90°$.

Therefore, it can be obtained that:

$$\begin{gathered} ∠NDA=∠NDC+∠CDA \\ =∠NDC+90° \end{gathered}$$

As, $∠FCEâ‰\dots ∠NDC$, therefore it can be obtained that:

$$\begin{gathered} ∠FCN=∠FCE+90° \\ =∠NDC+90° \end{gathered}$$

Therefore, $∠NDA=∠FCN$.

In the triangles $\mathrm{Δ}FCN$and $\mathrm{Δ}NDA$, it can be noticed that:

$FCâ‰\dots ND,∠NDAâ‰\dots ∠FCN$and$CNâ‰\dots DA$

Therefore, the triangles$\mathrm{Δ}FCN$ and$\mathrm{Δ}NDA$ are the congruent triangles by using the SAS postulate.

Therefore by using the corresponding parts of congruent triangles, it can be said that:

$FNâ‰\dots NA$.

As,$AFâ‰\dots NA$ and $FNâ‰\dots NA$, therefore it can be noticed that $AFâ‰\dots NAâ‰\dots FN$.

As, $AFâ‰\dots NAâ‰\dots FN$, therefore $AF=NA=FN$, therefore the triangle $\mathrm{Δ}FAN$is an equilateral triangle.

The two-column proof is: