Who is asking: Student
Question: how can you prove a quadrilateral to be a parallelogram?Hi Joy,
There are dozens of ways to 'prove' a quadrilateral is a parallelogram. Dozens of pairs of properties which are equivalent, in the sense that knowing one set gives each of the others.
Standard examples would be:
a quadrialteral with opposite sides equal (and not self-intersecting).
With so many to choose from, there is a single 'picture' that captures all the key facts: a quadrilateral ABCD which has a half-turn symmetry taking A to C, B to D, .... .
If you SEE this half turn symmetry, then all the samples I listed above become obvious properties.
If you take any to the list above (and many other pairs) then you can look for ways to PROVE the half turn symmetry from the pair of facts.
Therefore the two triangles are congruent. What IS the congruence - the map that takes one onto the other? With the common side AC = CA being reversed, and assuming B and D are on opposite sides of this common side, ( consequence of self- intersecting) the congruence IS a half-turn about the middle of AC.
Therefore the quadrilateral has all the other required properties, including opposite sides parallel, opposite angles parallel, etc.
So the approach I suggest, for many of the 'special quadrilaterals, is "look for the key symmetries". There is an analogous definition of a 'Kite" - a quadrilateral with a mirror symmetry in one diagonal. A rhombus is a quadrilateral with two lines of symmetry in the two diagonals. (You can move from these two mirrors to a half-turn, so a rhombus is a special parallelogram.)