Chapter 4.5, Problem 22WE

To determine

To prove: that triangle $\text{Δ}FAN$ is an equilateral triangle.

Answer to Problem 22WE

The triangle $\text{Δ}FAN$ is an equilateral triangle.

Explanation of Solution

Given information:

Triangle is an equilateral triangle.

System information: $\overline{AE}\cong \overline{FC}\cong \overline{ND}$

Calculation:

Consider the three squares ABCD, DEFG and CEMN.

And also one equilateral triangle DEC

Since all squares and triangle are joined to each other.

So it can be said that sides of all squares and triangle are equal.

Since $\text{Δ}ADE\approx \text{Δ}DCN$ implies that

  $AE=DN$

Also since $\angle DEA\approx \angle CDN$

To prove that triangle $\text{Δ}FAE$ is congruent to triangle $\text{Δ}DAN$

Consider triangle FAE & triangle DAN

Since $AD$ and $FE$ are sides of square and triangle so,

  $AD=FE$

Also as $AE=DN$

Consider the following,

  $\angle FEA=90+\angle DEA$

  $\angle ADN=90+\angle CDN$

So from above,

  $\angle \text{FEA}=\angle ADN$

Thus by SAS congruence criteria $\text{Δ}FAE\approx \text{Δ}DAN$

Thus it can be said that $\text{FA}=AN$

Similarly consider that $\text{Δ}FEC\approx \text{Δ}DCN$ implies that

  $FC=DN$

Also as $\angle ECF=\angle CDN$

To prove that triangle $FCN$ is congruent to triangle $DAN$

Consider triangle $CN$ and triangle DAN

Since $AD=CN$ and $CN$ are sides of square and triangle so,

  $AD=CN$

Also as $FC=DN$

Consider the following,

  $\angle FCN=90+\angle ECF$

  $\angle ADN=90+\angle CDN$

So from above,

  $\angle \text{FCN}=\angle ADN$

Thus by SAS congruence criteria $\text{Δ}FCN\approx \text{Δ}DAN$

Thus it can be said that $\text{FN}=AN$

Consider the sides as,

  $FA=AN=FN$

Therefore, triangle $\text{Δ}FAN$ is an equilateral triangle.