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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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