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Let O be a point inside the quadrilateral ABCD.

AOB+COB+COD+DOA=360° because the angles at the centre must sum to a complete circle.

AOB=180-(OAB+ABO), COB=180-(OBC+CBO), COD=180-(ODC+OCD), DOA=180-(ODA+OAD).

If we add these together we get:

720-(DAB+ABC+DCB+ADC)=360, because OAB+DAO=DAB, etc.

Another way of writing this is to ignore the vertex connections to O, so:

720-A+B+C+D=360, that is, A+B+C+D=360, meaning the interior angles of a quadrilateral sum to 360°.

But we are given C=A and D=B, so 2A+2B=360 and A+B=180, so A and B are supplementary angles, and therefore C and D are supplementary.

Two angles on a straight line are supplementary. We now extend AB on both sides to P and Q so that PABQ are colinear and similarly extend DC to RDCS. D=B=180-A. CBQ=180-(180-A)=DAB=A. Therefore AD and BC must have the same slope, because DAB and CBQ are congruent and are corresponding angles, so AD and BC are parallel. But D=180-C=180-A given A=C. Since A and D are supplementary because A=C we have all the angles of the quadrilateral in terms of A: B=D=180-A, C=A. BCS=ABC(=B)=ADC(=D)=180-A. This proves that AB and DC are parallel. The quadrilateral has two pairs of parallel sides, so it’s a parallelogram.

 

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