Given: A quadrilateral ABCD in a graph.
To Prove: ABCD is a parallelogram.
Construction: Draw a diagonal AC. Now 2 triangles ACD and ACB are formed.
Proof:
As the figure shows, the sides AB and CD are 4 cm.
(A is at -2 and B is at 2. If we add 2 cm to both sides of the graph, we get 4 cm. Similarly,
CD extends from 0 to -4 which makes it 4 cm in length.
Hence, AB=DC ...(i)
Take the angles <BAC and <ACD. As they are alternate interior angles (As AB=CD and AB || CD as visible from the figure),
<BAC = <ACD …(ii)
It can also be noted that the triangles ACD and ACB have a common base AC.
Hence, AC=AC …(iii)
From equations i, ii, and iii, ACB ≡ ACD (Congruent) by SAS congruence.
So, the pair of opposite sides AD and BC are equal due to their congruence. Hence, if the BD diagonal is constructed, it can be proven that the triangles BDA and BDC in a similar way.
Hence it is proven that the pair of opposite sides are equal and parallel. (Sides cannot be congruent, only triangles can). So, the quadrilateral ABCD is a parallelogram.
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