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It is given that ABCD is a trapezium with AB parallel to CD.

Draw DE and CF perpendicular to AB.

In triangle ABC, angle BAC is acute, so we have:

BC² = AC² + AB² - 2AF. AB              ... (1)

In triangle BDA angle DBA is acute, so we have:

AD² = BD² + AB² - 2BE. AB             ... (1)

Adding (1) and (2), we have:

BC² + AD² = AC² + BD² + 2AB² - 2AF. AB - 2BE. AB

BC² + AD² = AC² + BD² + 2AB[AB - AF - BE]

BC² + AD² = AC² + BD² + 2AB[AB - (AE + EF) - (BF + EF)]

BC² + AD² = AC² + BD² + 2AB[AB - (AE + EF + BF + EF)]

BC² + AD² = AC² + BD² + 2AB[AB - (AB + CD)]

BC² + AD² = AC² + BD² - 2AB. CD

AC² + BD² = BC² + AD² + 2AB. CD