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In-a-quadrilateral-ABCD-it-is-given-that-AB-is-parallel-to-CD-and-the-diagonals-AC-and-BD-are-perpendicular-to-each-other-Show-that-a-AD-BC-AB-CD-b-AD-BC-AB-CD-




Question Number 22515 by Tinkutara last updated on 19/Oct/17
In a quadrilateral ABCD, it is given  that AB is parallel to CD and the  diagonals AC and BD are perpendicular  to each other.  Show that  (a) AD.BC ≥ AB.CD;  (b) AD + BC ≥ AB + CD.
$$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:{ABCD},\:\mathrm{it}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{that}\:{AB}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{CD}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{diagonals}\:{AC}\:\mathrm{and}\:{BD}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{AD}.{BC}\:\geqslant\:{AB}.{CD}; \\ $$$$\left(\mathrm{b}\right)\:{AD}\:+\:{BC}\:\geqslant\:{AB}\:+\:{CD}. \\ $$

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