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ABCD is a quadrilateral in which all four sides are equal. Show that both pairs of opposite sides are equal.

Answer
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Hint: Here, we will first draw quadrilateral ABCD with same sides length. Then we will draw AC and BC bisectors which will divide angles in equal parts. So, by this we will prove that BAC=DAC=ACD=BCA and ABD=CBD=ADB=CDB . Then we will use the concept of alternate angle theorem which is given as two angles formed when a line crosses two other lines that lie on opposite sides of the transversal lines and opposite relative sides of the other lines. Then by converse theorem we will prove both pairs of opposite sides are equal.

Complete step-by-step answer:
Here, if we assume that sides of quadrilateral ABCD are the same then interior angles will be also same with 90 degrees. So, we get figure as
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Thus, we can say that ABCD is square.
Now, if we draw diagonals AC and BD then, they will bisect interior angles and will meet at point O.
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Now, we can say BAD is bisected by AC and BCD is bisected by AC.
So, we can say that ACD=ACB and BAC=DAC .
Now, since BAD=BCD=90 , we can say that BAC=DAC=ACD=BCA .
By using alternate angle theorem which is given as two angles formed when a line crosses two other lines that lie on opposite sides of the transversal lines and opposite relative sides of the other lines.
We can say that BAC=ACD .
So, by converse theorem we can say that AB is parallel to CD. ……………..(1)
Similarly, if we take BD as bisector, we get angles to be equal as ABD=CBD=ADB=CDB
Thus, we can say that CBD=ADB by alternate angles theorem.
So, by converse theorem we can say that AD is parallel to BC. …………………..(2)
Hence, opposite sides are parallel quadrilateral ABCD are equal. Thus, proved.

Note: Students should know the concept of bisector. This is not to be proved by saying that as this quadrilateral is square then we know that the length of opposite sides is the same. We have to prove the interior angles are equal and then by converse theorem we have to give an answer. So, be careful while solving this type of problem.