Finding the diagonal of an irregular quadrilateral

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To find the diagonal AC in an irregular quadrilateral ABCD, where the sides are labeled AB=a, BC=b, CD=c, and DA=d, the relationship between the sides and the diagonal can be established using the cosine formula. The equation to prove is (ab+cd)x^2=(ac+bd)(ad+bc), with the condition that angle ABC + angle ADC = 180 degrees. The cosine formula can be applied twice, first for triangle ABC and then for triangle ADC, treating AC as the diagonal x. By setting the two expressions for x equal, the proof can be developed. This approach effectively utilizes the properties of triangles and the cosine law to establish the required relationship.
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Homework Statement


In an irregular quadrilateral ABCD, the length of all sides are AB=a BC=b CD=c DA=d and the length of the diagonal AC is x. Angle ABC + angle ADC = 180

prove that (ab+cd)x2=(ac+bd)(ad+bc)

Homework Equations



Cosine formula c2 = a2 + b2 - 2abcosθ

The Attempt at a Solution


I really have no idea how to start
 
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Use the cosine formula, c^2= a^2+ b^2- 2abcos(C), with x as "c", sides AB and BC as "a" and "b", and angle ABC as angle "C". Then do the same with x as "c" again but AD and BD as "a" and "b", and angle ADC as angle "C". Of course, the "x" in both of those is the same so they can be set equal.
 

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