Recall that in a parallelogram diagonals bisect each other.
See solution.
Practice makes perfect
Let's take a look at the given picture and draw the second diagonal of parallelogram ABCD. Let P be the point of intersection of the diagonals of ABCD. Since ABCD is a parallelogram, its diagonals bisect each other.
Now, we have a quadrilateral JBKD that lies inside this parallelogram and we know that AJ=KC. Notice that it also has a diagonal of BD.
From the Properties of Parallelograms, we know that AP=PC. Let's use the Segment Addition Postulate and rewrite each of these diagonal as a sum of two segments. Remember that we are given that AJ=KC.
We got that JP=PK. This means that the diagonals of JBKD also bisect each other.
From the Conditions of Parallelograms, we know that if the diagonals of a quadrilateral bisect each other, then this quadrilateral is a parallelogram. Therefore, quadrilateral JBKD is a parallelogram.
