Prove that the sum of the diagonals of a trapezoid is greater than the sum of its bases.

Consider triangles AOD and BCO formed by intersecting diagonals and trapezoid bases.

Because each side of a triangle is less than the sum of the other two sides, then:

AD <AO + OD (for triangle AOD) and BC <BO + OC (for triangle BCO).

Let us add up the inequalities.

AD + BC <AO + OD + BO + OC;

AD + BC <(AO + OC) + (OD + BO);

AD + BC <AC + BD.

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