Answer
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- Hint: For solving this question, we will consider a regular hexagon ABCDEF and then we will join two pairs of points. After that, we will use some of the properties like for regular hexagon length of all sides will be equal and all internal angles will be equal to ${{120}^{\circ }}$ , and then we will use the congruence rule Side Angle Side to proceed further. Then, we will make use of the properties of the rectangle and triangle together to prove the desired result easily.
Complete step-by-step solution -
Given:
We have a regular hexagon and we have to prove that its opposite sides are parallel to each other.
Now, let there be a regular hexagon ABCDEF. For more clarity, look at the figure given below:
In the above figure, ABCDEF is a regular hexagon, which means the length of all sides will be equal and all internal angles will be equal to ${{120}^{\circ }}$. Then,
$\begin{align}
& AB=BC=CD=DE=EF=FA.................\left( 1 \right) \\
& \angle FAB=\angle ABC=\angle BCD=\angle CDE=\angle DEF=\angle EFA={{120}^{\circ }}................\left( 2 \right) \\
\end{align}$
To Prove: We have to prove that, $AB\parallel DE$ , $BC\parallel EF$ and $CD\parallel FA$ .
Now, we do construction and join points $F\And B$ by segment FB and points $E\And C$ by segment EC. For more clarity look at the figure given below:
In the above figure, we will consider $\Delta FAB$ and $\Delta CDE$ .
Now, in $\Delta FAB$ and $\Delta CDE$ . We have:
1. $FA=CD$ (from equation (1)).
2. $AB=DE$ (from equation (1)).
3. $\angle FAB=\angle CDE$ (from equation (2)).
Now, by considering the above 3 points with the help of SAS (Side Angle Side) congruence rule we conclude that: $\Delta FAB\cong \Delta CDE$ . As corresponding sides of congruent triangles are equal in length. Then,
$FB=CE...................\left( 3 \right)$
Now, as we know the sum of interior angles of a triangle is always ${{180}^{0}}$ and angles opposite to equal sides of an isosceles triangle are equal.
Now, as in $\Delta FAB$ & $\Delta CDE$ we have $\angle FAB=\angle CDE={{120}^{\circ }}$ . Then,
$\begin{align}
& \angle FAB+\angle AFB+\angle ABF={{180}^{\circ }} \\
& \Rightarrow {{120}^{\circ }}+\angle AFB+\angle ABF={{180}^{\circ }} \\
& \Rightarrow \angle AFB+\angle ABF={{60}^{\circ }}.................\left( 4 \right) \\
& \angle CDE+\angle DCE+\angle DEC={{180}^{\circ }} \\
& \Rightarrow {{120}^{\circ }}+\angle DCE+\angle DEC={{180}^{\circ }} \\
& \Rightarrow \angle DCE+\angle DEC={{60}^{\circ }}.................\left( 5 \right) \\
\end{align}$
Moreover, in both triangles, we have a pair of equal sides as $FA=AB$ and $CD=DE$ which means they are isosceles triangles. So, by the property of the isosceles triangle, we can put $\angle ABF=\angle AFB$ in equation (4) and $\angle DCE=\angle DEC$ in equation (5). Then,
\[\begin{align}
& \angle AFB+\angle ABF={{60}^{\circ }} \\
& \Rightarrow \angle AFB=\angle ABF={{30}^{\circ }}.............\left( 6 \right) \\
& \angle DCE+\angle DEC={{60}^{\circ }} \\
& \Rightarrow \angle DCE=\angle DEC={{30}^{\circ }}.............\left( 7 \right) \\
\end{align}\]
Now, consider the figure below:
Then,
$\begin{align}
& \angle EFA=\angle EFB+\angle AFB={{120}^{\circ }} \\
& \angle DEF=\angle FEC+\angle DEC={{120}^{\circ }} \\
& \angle ABC=\angle FBC+\angle ABF={{120}^{\circ }} \\
& \angle BCD=\angle BCE+\angle DCE={{120}^{\circ }} \\
\end{align}$
Now, we can put \[\angle AFB=\angle ABF={{30}^{\circ }}\] from equation (6) and \[\angle DCE=\angle DEC={{30}^{\circ }}\] from equation (7) in the above equations. Then,
$\begin{align}
& \angle EFA=\angle EFB+\angle AFB={{120}^{\circ }} \\
& \Rightarrow \angle EFB+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle EFB={{90}^{\circ }}.........................\left( 8 \right) \\
& \angle DEF=\angle FEC+\angle DEC={{120}^{\circ }} \\
& \Rightarrow \angle FEC+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle FEC={{90}^{\circ }}.........................\left( 9 \right) \\
& \angle ABC=\angle FBC+\angle ABF={{120}^{\circ }} \\
& \Rightarrow \angle FBC+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle FBC={{90}^{\circ }}.........................\left( 10 \right) \\
& \angle BCD=\angle BCE+\angle DCE={{120}^{\circ }} \\
& \Rightarrow \angle BCE+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle BCE={{90}^{\circ }}.........................\left( 11 \right) \\
\end{align}$
Now, consider the quadrilateral $EFBC$ . We have:
1. $EF=BC$ (from equation (1)).
2. $FB=CE$ (from equation (3)).
3. $\angle EFB=\angle FEC=\angle FBC=\angle BCE={{90}^{\circ }}$ (from equations (8), (9), (10) and (11)).
Now, from the above result, we conclude that quadrilateral $EFBC$ is a rectangle and we know that, in a rectangle opposite sides are parallel. Then,
$BC\parallel EF$
Now, similarly, we can write that $AB\parallel DE$ , and $CD\parallel FA$ .
Thus, opposite sides of a regular hexagon are parallel.
Hence, proved.
Note: Here, the student must take care of the geometrical properties of the regular hexagon. And we should be careful in writing the name of congruent triangles. For example in this question we cannot write that $\Delta FBA\cong \Delta CDE$ , the correct way is $\Delta FAB\cong \Delta CDE$ . Moreover, we should make use of every geometrical property like the sum of interior angles of triangles and make use of the property of rectangles accurately without any mathematical error to prove the desired result easily. And we should remember this result.
Complete step-by-step solution -
Given:
We have a regular hexagon and we have to prove that its opposite sides are parallel to each other.
Now, let there be a regular hexagon ABCDEF. For more clarity, look at the figure given below:
In the above figure, ABCDEF is a regular hexagon, which means the length of all sides will be equal and all internal angles will be equal to ${{120}^{\circ }}$. Then,
$\begin{align}
& AB=BC=CD=DE=EF=FA.................\left( 1 \right) \\
& \angle FAB=\angle ABC=\angle BCD=\angle CDE=\angle DEF=\angle EFA={{120}^{\circ }}................\left( 2 \right) \\
\end{align}$
To Prove: We have to prove that, $AB\parallel DE$ , $BC\parallel EF$ and $CD\parallel FA$ .
Now, we do construction and join points $F\And B$ by segment FB and points $E\And C$ by segment EC. For more clarity look at the figure given below:
In the above figure, we will consider $\Delta FAB$ and $\Delta CDE$ .
Now, in $\Delta FAB$ and $\Delta CDE$ . We have:
1. $FA=CD$ (from equation (1)).
2. $AB=DE$ (from equation (1)).
3. $\angle FAB=\angle CDE$ (from equation (2)).
Now, by considering the above 3 points with the help of SAS (Side Angle Side) congruence rule we conclude that: $\Delta FAB\cong \Delta CDE$ . As corresponding sides of congruent triangles are equal in length. Then,
$FB=CE...................\left( 3 \right)$
Now, as we know the sum of interior angles of a triangle is always ${{180}^{0}}$ and angles opposite to equal sides of an isosceles triangle are equal.
Now, as in $\Delta FAB$ & $\Delta CDE$ we have $\angle FAB=\angle CDE={{120}^{\circ }}$ . Then,
$\begin{align}
& \angle FAB+\angle AFB+\angle ABF={{180}^{\circ }} \\
& \Rightarrow {{120}^{\circ }}+\angle AFB+\angle ABF={{180}^{\circ }} \\
& \Rightarrow \angle AFB+\angle ABF={{60}^{\circ }}.................\left( 4 \right) \\
& \angle CDE+\angle DCE+\angle DEC={{180}^{\circ }} \\
& \Rightarrow {{120}^{\circ }}+\angle DCE+\angle DEC={{180}^{\circ }} \\
& \Rightarrow \angle DCE+\angle DEC={{60}^{\circ }}.................\left( 5 \right) \\
\end{align}$
Moreover, in both triangles, we have a pair of equal sides as $FA=AB$ and $CD=DE$ which means they are isosceles triangles. So, by the property of the isosceles triangle, we can put $\angle ABF=\angle AFB$ in equation (4) and $\angle DCE=\angle DEC$ in equation (5). Then,
\[\begin{align}
& \angle AFB+\angle ABF={{60}^{\circ }} \\
& \Rightarrow \angle AFB=\angle ABF={{30}^{\circ }}.............\left( 6 \right) \\
& \angle DCE+\angle DEC={{60}^{\circ }} \\
& \Rightarrow \angle DCE=\angle DEC={{30}^{\circ }}.............\left( 7 \right) \\
\end{align}\]
Now, consider the figure below:
Then,
$\begin{align}
& \angle EFA=\angle EFB+\angle AFB={{120}^{\circ }} \\
& \angle DEF=\angle FEC+\angle DEC={{120}^{\circ }} \\
& \angle ABC=\angle FBC+\angle ABF={{120}^{\circ }} \\
& \angle BCD=\angle BCE+\angle DCE={{120}^{\circ }} \\
\end{align}$
Now, we can put \[\angle AFB=\angle ABF={{30}^{\circ }}\] from equation (6) and \[\angle DCE=\angle DEC={{30}^{\circ }}\] from equation (7) in the above equations. Then,
$\begin{align}
& \angle EFA=\angle EFB+\angle AFB={{120}^{\circ }} \\
& \Rightarrow \angle EFB+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle EFB={{90}^{\circ }}.........................\left( 8 \right) \\
& \angle DEF=\angle FEC+\angle DEC={{120}^{\circ }} \\
& \Rightarrow \angle FEC+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle FEC={{90}^{\circ }}.........................\left( 9 \right) \\
& \angle ABC=\angle FBC+\angle ABF={{120}^{\circ }} \\
& \Rightarrow \angle FBC+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle FBC={{90}^{\circ }}.........................\left( 10 \right) \\
& \angle BCD=\angle BCE+\angle DCE={{120}^{\circ }} \\
& \Rightarrow \angle BCE+{{30}^{\circ }}={{120}^{\circ }} \\
& \Rightarrow \angle BCE={{90}^{\circ }}.........................\left( 11 \right) \\
\end{align}$
Now, consider the quadrilateral $EFBC$ . We have:
1. $EF=BC$ (from equation (1)).
2. $FB=CE$ (from equation (3)).
3. $\angle EFB=\angle FEC=\angle FBC=\angle BCE={{90}^{\circ }}$ (from equations (8), (9), (10) and (11)).
Now, from the above result, we conclude that quadrilateral $EFBC$ is a rectangle and we know that, in a rectangle opposite sides are parallel. Then,
$BC\parallel EF$
Now, similarly, we can write that $AB\parallel DE$ , and $CD\parallel FA$ .
Thus, opposite sides of a regular hexagon are parallel.
Hence, proved.
Note: Here, the student must take care of the geometrical properties of the regular hexagon. And we should be careful in writing the name of congruent triangles. For example in this question we cannot write that $\Delta FBA\cong \Delta CDE$ , the correct way is $\Delta FAB\cong \Delta CDE$ . Moreover, we should make use of every geometrical property like the sum of interior angles of triangles and make use of the property of rectangles accurately without any mathematical error to prove the desired result easily. And we should remember this result.
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